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Bank Soal Matematika Dasar Eksponen (*Soal dan Pembahasan)

LogaritmaKesulitan menganalisa kalimat soal$a^{m}= \underset{perkalian\ sebanyak\ m}{\underbrace{a \cdot a \cdot a \cdot a \cdots a \cdot a}}$
$m:$ Bilangan pangkat [Eksponen]
$a:$ Bilangan Pokok [Basis]
$0^{0}=$ tidak terdefenisi

$a^{m} \cdot a^{n}=a^{m+n}$

$\dfrac{a^{m}}{a^{n}}=a^{m-n}$

$(a^{m})^{n}=a^{m \cdot n}$

$a^{m} \cdot b^{m}=(a \cdot b)^{m}$

$\dfrac{a^{m}}{b^{m}} = \left( \dfrac{a}{b} \right )^{m}$

$\dfrac{1}{a^{m}}={a}^{-m}$ dengan $a \neq 0$

$\dfrac{1}{a^{-m}}={a}^{m}$ dengan $a \neq 0$

$a^{0}=1$ dengan $a \neq 0$

$a^{\frac{m}{n}}=\sqrt[n]{a^{m}}$

Jika $a^{f(x)}=a^{g(x)}$ maka $f(x)=g(x)$

1. Soal SNMPTN 2010 Kode 336 (*Soal Lengkap)Jika $n$ memenuhi $\underset{n\ faktor}{\underbrace{25^{0.25} \times 25^{0.25}\times \cdots \times 25^{0.25}\times 25^{0.25}}=125}$
maka $(n-3)(n+2)=\cdots$
$\begin{align}
(A)\ &24 \\
(B)\ &26 \\
(C)\ &28 \\
(D)\ &32 \\
(E)\ &36
\end{align}$Alternatif Pembahasan: show

$\begin{align}
25^{0.25} \times 25^{0.25}\times \cdots \times 25^{0.25}\times 25^{0.25} &= 125 \\
5^{0.5} \times 25^{0.5}\times \cdots \times 5^{0.5}\times 25^{0.5} &= 5^{3} \\
5^{0.5} \times 5^{0.5}\times \cdots \times 5^{0.5}\times 5^{0.5} &= 5^{3} \\
\left(5^{0.5}\right)^{n} &= 5^{3} \\
5^{\dfrac{1}{2}n} &= 5^{3} \\
0.5n &= 3 \\
n &=6 \\
(n-3)(n+2) &= (6-3)(6+2) \\
&=24
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(A)\ 24$

2. Soal SIMAK UI 2009 Kode 951 (*Soal lengkap)Diketahui $x_{0}$ dan $y_{0}$ adalah nilai-nilai yang memenuhi sistem persamaan $\begin{cases}2^{x+1}-3^{y}=7 \\ -\left(2^{x-1} \right)-3^{y+1}=-5\end{cases}$
maka $x_{0}+y_{0}$ adalah$\cdots$
$\begin{align}
(A)\ & -1 \\
(B)\ & 0 \\
(C)\ & 1 \\
(D)\ & 1 \\
(E)\ & 2
\end{align}$Alternatif Pembahasan: show

$\begin{align}
2^{x+1}-3^{y} &= 7 \\
2^{x} \cdot 2^{1}-3^{y} &= 7 \\
2^{x} \cdot 2-3^{y} &= 7
\end{align}$

$\begin{align}
-\left(2^{x-1} \right)-3^{y+1} &= -5 \\
2^{x-1}+3^{y+1} &= 5 \\
2^{x} \cdot 2^{-1}+3^{y} \cdot 3^{1} &= 5 \\
2^{x} \cdot \dfrac{1}{2}+3^{y} \cdot 3 &= 5 \\
2^{x} +3^{y} \cdot 6 &= 10
\end{align}$

Dengan memisalkan $m=2^{x}$ dan $n=3^{y}$, maka sistem persamaan dapat kita ubah sementara menjadi;
$\begin{array}{c|c|cc}
2m-n = 7 & (\times 1) \\
m+6n = 10 & (\times 2) \\
\hline
2m-n = 7 & \\
2m+12n = 20 & (-) \\
\hline
-13n = -13 & 2m-1 = 7 \\
n = 1 & m = 4
\end{array} $

$m=2^{x}$ $\Rightarrow$ $4=2^{x}$ $\Rightarrow$ $x=2$
$n=3^{y}$ $\Rightarrow$ $1=3^{y}$ $\Rightarrow$ $y=0$
Nilai $x_{0}+y_{0}=2+0=2$

$\therefore$ Pilihan yang sesuai adalah $(D)\ 2$

3. Soal SPMB 2003 [Regional I] (*Soal Lengkap)Nilai $x$ yang memenuhi persamaan $3^{2x+3}=\sqrt[3]{27^{x+5}}$ adalah$\cdots$
$\begin{align}
(A)\ & -2 \\
(B)\ & -1 \\
(C)\ & 0 \\
(D)\ & 1 \\
(E)\ & 2
\end{align}$Alternatif Pembahasan: show

$\begin{split}
3^{2x+3} &=\sqrt[3]{27^{x+5}}\\
3^{2x+3} &=27^{\dfrac{x+5}{3}}\\
3^{2x+3} &=(3^{3})^{\dfrac{x+5}{3}}\\
3^{2x+3} &=3^{x+5}\\
& \Rightarrow 2x+3=x+5\\
& \Rightarrow 2x-x=5-3\\
& \Rightarrow x=2
\end{split}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 2$

4. Soal SIMAK UI 2012 Kode 221 (*Soal lengkap)Jika diketahui $x$ dan $y$ adalah bilangan real dengan $x \gt 1$ dan $y \gt 0$. Jika $xy=x^{y}$ dan $\dfrac{x}{y}=x^{5y}$, maka $x^{2}+3y=\cdots$
$\begin{align}
(A)\ & 29 \\
(B)\ & 28 \\
(C)\ & 27 \\
(D)\ & 26 \\
(E)\ & 25
\end{align}$Alternatif Pembahasan: show

$\begin{align}
xy &= x^{y} \\
y &= \dfrac{x^{y}}{x} \\
y &= x^{y-1}
\end{align}$

$\begin{align}
\dfrac{x}{y} &= x^{5y} \\
\dfrac{x}{x^{y-1}} &= x^{5y} \\
x &= x^{5y} \cdot x^{y-1} \\
x &= x^{6y-1} \\
& \Rightarrow 1=6y-1 \\
& \Rightarrow 2=6y \\
& \Rightarrow y=\dfrac{1}{3}
\end{align}$

Jika kita substitusikan pers.(1) dan pers.(2) maka kita peroleh;
$\begin{align}
y-1 &= 1-5y \\
6y &= 2 \\
y &= \dfrac{1}{3}
\end{align}$

$\begin{align}
xy &= x^{y} \\
x \cdot \frac{1}{3} &= x^{\frac{1}{3}} \\
x &= 3 x^{\frac{1}{3}} \\
x \cdot x^{-\frac{1}{3}} &= 3 \\
x^{ \frac{2}{3}} &= 3 \\
x^{2} &= 3^{3} \\
x^{2}+3y &= 3^{3} + 3 \cdot \frac{1}{3} = 28
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 28$

5. Soal SPMB 2005 Kode 470 (*Soal Lengkap)Jika $f(x)=2^{2x}+2^{x+1}-3$ dan $g(x)=2^{x}+3$ maka $\dfrac{f(x)}{g(x)}=\cdots$
$\begin{align}
(A)\ & 2^{x}+3 \\
(B)\ & 2^{x}+1 \\
(C)\ & 2^{x} \\
(D)\ & 2^{x}-1 \\
(E)\ & 2^{x}-3
\end{align}$Alternatif Pembahasan: show

$\begin{align}
\dfrac{f(x)}{g(x)} &= \dfrac{2^{2x}+2^{x+1}-3}{2^{x}+3} \\
&=\dfrac{(2^{x})^{2}+2^{x} \cdot 2^{1}-3}{2^{x}+3} \\
&=\dfrac{(2^{x})^{2}+2^{x} \cdot 2^{1}-3}{2^{x}+3}
\end{align}$

Untuk mempermudah penglihatan, mungkin $2^{x}$ sementara bisa kita ganti menjadi $m$.
$\begin{align}
\dfrac{f(x)}{g(x)} &= \dfrac{(m)^{2}+ m \cdot 2^{1}-3}{m+3} \\
&= \dfrac{m^{2}+2m-3}{m+3} \\
&= \dfrac{(m+3)(m-1)}{m+3} \\
&= m-1 \\
&= 2^{x}-1
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ 2^{x}$-1

6. Soal SIMAK UI 2013 Kode 437 (*Soal lengkap)Diketahui bahwa $2^{w} \cdot a^{x} \cdot b^{y} \cdot c^{z}=2013$ untuk setiap $a,b,c,d,x,y,z$ merupakan bilangan bulat positif dan $w$ bilangan bulat nonnegative dengan $a \lt b \lt c$. Nilai $(2w)+(ax)+(by)+(cz)=\ldots$
$\begin{align}
(A)\ & 0 \\
(B)\ & 3 \\
(C)\ & 11 \\
(D)\ & 75 \\
(E)\ & 611
\end{align}$Alternatif Pembahasan: show

$\begin{align}
2^{w} \cdot a^{x} \cdot b^{y} \cdot c^{z} &= 2013 \\
2^{w} \cdot a^{x} \cdot b^{y} \cdot c^{z} &= 3 \cdot 11 \cdot 61 \\
2^{w} \cdot a^{x} \cdot b^{y} \cdot c^{z} &= 2^{0} \cdot 3^{1} \cdot 11^{1} \cdot 61^{1}
\end{align}$
Sehingga diperoleh; $w=0$, $x=1$, $y=1$, $z=1$, $a=3$, $b=11$, $c=61$

$\begin{align}
&(2w)+(ax)+(by)+(cz) \\
&= (2 \cdot 0)+(3 \cdot 1)+(11 \cdot 1)+(61 \cdot 1) \\
&= 0+3+11+61 \\
&= 75
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ 2^{x}$-1

7. Soal UM UGM 2017 Kode 814 (*Soal Lengkap)Jika $f(x)=b^{x}$, $b$ konstanta positif, maka $\dfrac{f(x^{2}-1)}{f(1-x^{2})}=\cdots$
$\begin{align}
(A)\ & f(1-x^{2}) \cdot f(1-x^{2}) \\
(B)\ & f(1-x^{2}) \cdot f(x^{2}-1) \\
(C)\ & f(x^{2}-1) \cdot f(x^{2}-1) \\
(D)\ & f(1-x^{2}) + f(1-x^{2}) \\
(E)\ & f(x^{2}-1) + f(x^{2}-1)
\end{align}$Alternatif Pembahasan: show

$\begin{align}
& \dfrac{f(x^{2}-1)}{f(1-x^{2})} = \dfrac{b^{x^{2}-1}}{b^{1-x^{2}}} \\
&= \dfrac{b^{x^{2}} \cdot b^{-1}}{b^{1} \cdot b^{-x^{2}}} = \dfrac{b^{x^{2}} \cdot b^{x^{2}}}{b^{1} \cdot b^{1}} \\
&= \dfrac{b^{2x^{2}}}{b^{2}} = b^{2x^{2}-2} \\
&= b^{2(x^{2}-1)} = \left(b^{x^{2}-1} \right)^2 \\
&= \left(b^{x^{2}-1} \right) \cdot \left(b^{x^{2}-1} \right) \\
&= f(x^{2}-1) \cdot f(x^{2}-1)
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ f(x^{2}-1) \cdot f(x^{2}-1)$

8. Soal SIMAK UI 2014 Kode 511 (*Soal Lengkap)Dalam basis 10, bilangan bulat positif $p$ memiliki $3$ digit, bilangan bulat positif $q$ memiliki $p$ digit, bilangan bulat positif $r$ memiliki $q$ digit. Nilai untuk terkecil untuk $r$ adalah$\cdots$
$\begin{align}
(A)\ & 10^{10^{100}} \\
(B)\ & 10^{10^{100}-1} \\
(C)\ & 10^{10^{99}} \\
(D)\ & 10^{10^{99}-1} \\
(E)\ & 10^{99^{99}}
\end{align}$Alternatif Pembahasan: show

Topik ini sebenarnya tidak murni tentang eksponen, tetapi karena pilihannya bilangan berpangkat para siswa melihat ini tentang bilangan berpangkat. Ada sedikit logika atau teori bilangan didalamnya.

Pada soal diinginkan agar nilai bilangan $r$ mempunyai nilai terkecil, maka bilangan $q$ kita juga harus bilangan terkecil. Sehingga bilangan $p$ juga harus memiliki nilai terkecil.

Bilangan $p$ terdiri dari $3$ digit, supaya mendapatkan $p$ bilangan terkecil maka angka pertama [ratusan] dipilih angka $1$ dan sisanya [puluhan dan satuan] dipilih angka nol sehingga $p = 100= 10^{3-1} = 10^{2}$

Bilangan $q$ terdiri dari $100$ digit, supaya mendapatkan $q$ bilangan terkecil maka angka pertama dipilih angka $1$ dan sisanya dipilih angka nol sehingga $q = 10^{100-1} = 10^{99}$

Bilangan $r$ terdiri dari $q$ digit, supaya mendapatkan $r$ bilangan terkecil maka angka pertama dipilih angka $1$ dan sisanya dipilih angka nol sehingga $r = 10^{10^{99}-1}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ 10^{10^{99}-1}$

9. Soal UM UGM 2005 Kode 821 (*Soal Lengkap)Nilai $x$ yang memenuhi $\dfrac{2^{x}}{4^{x+2}}=16 \cdot 4^{x}$ adalah$\cdots$
$\begin{align}
(A)\ & -3 \\
(B)\ & -\dfrac{8}{3} \\
(C)\ & -2 \\
(D)\ & -\dfrac{4}{3} \\
(E)\ & -\dfrac{2}{3}
\end{align}$Alternatif Pembahasan: show

$\begin{align}
\dfrac{2^{x}}{4^{x+2}} &=16 \cdot 4^{x} \\
2^{x} &=2^{4} \cdot 4^{x} \cdot 4^{x+2} \\
2^{x} &=2^{4} \cdot 2^{2x} \cdot 2^{2x+4} \\
2^{x} &=2^{4+2x+2x+4} \\
2^{x} &=2^{4x+8} \\
x &=4x+8 \\
-3x &=8 \\
x &=-\dfrac{8}{3}
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ -\dfrac{8}{3}$

10. Soal SIMAK UI 2015 Kode 563 (*Soal Lengkap)$\dfrac{2015^{2}(2014^{2}-2013)}{(2014^{2}-1)(2014^{3}+1)}\times \dfrac{2013^{2}(2014^{2}+2015)}{(2014^{3}-1)}=\cdots$
$\begin{align}
(A)\ & 2013 \times 2015\\
(B)\ & 2015 \\
(C)\ & 2014 \\
(D)\ & 2013 \\
(E)\ & 1
\end{align}$Alternatif Pembahasan: show

Untuk mengerjakan soal ini agar penulisan dan pemfaktoran lebih mudah dioahami kita gunakan pemisalan, yaitu:
$m=2014$ sehingga $m-1=2013$ dan $m+1=2015$

$\begin{align}
& \dfrac{2015^{2}(2014^{2}-2013)}{(2014^{2}-1)(2014^{3}+1)}\times \dfrac{2013^{2}(2014^{2}+2015)}{(2014^{3}-1)}
&=\dfrac{(m+1)^{2}(m^{2}-(m-1))}{(m^{2}-1)(m^{3}+1)}\times \dfrac{(m-1)^{2}(m^{2}+(m+1))}{(m^{3}-1)} \\
&=\dfrac{(m+1)^{2}(m^{2}-m+1)}{(m^{2}-1)(m^{3}+1)}\times \dfrac{(m-1)^{2}(m^{2}+m+1)}{(m^{3}-1)} \\
&=\dfrac{(m+1)(m+1)(m^{2}-m+1)}{(m-1)(m+1)(m^{3}+1)}\times \dfrac{(m-1)(m-1)(m^{2}+m+1)}{(m^{3}-1)} \\
&=\dfrac{(m+1)(m+1)(m^{2}-m+1)(m-1)(m-1)(m^{2}+m+1)}{(m-1)(m+1)(m^{3}+1)(m^{3}-1)} \\
&=\dfrac{(m+1)(m^{2}-m+1)(m-1)(m^{2}+m+1)}{(m^{3}+1)(m^{3}-1)} \\
&=\dfrac{(m^{3}+1)(m^{3}-1)}{(m^{3}+1)(m^{3}-1)} \\
&=1
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 1$

11. Soal Matematika Piral (*Soal Lengkap)Nilai dari $\dfrac{1}{10^{-2017}+1}+\dfrac{1}{10^{-2016}+1}+\dfrac{1}{10^{-2015}+1}$$+\cdots+\dfrac{1}{10^{0}+1}+\cdots+$$\dfrac{1}{10^{2015}+1}+\dfrac{1}{10^{2016}+1}+\dfrac{1}{10^{2017}+1}$ adalah...
$\begin{align}
(A)\ & 2015,5\\
(B)\ & 2017,5 \\
(C)\ & 0 \\
(D)\ & 1 \\
(E)\ & 2017
\end{align}$Alternatif Pembahasan: show

Untuk mengerjakan soal ini jika kita kerjakan satu persatu pastinya akan melelahkan, karena penjumlahan pecahan sampai $2017$ kali, sehingga dibutuhkan kreatifitas, kita butuh pilar (pintar bernalar).

Kita coba dengan menjumlahkan yang kelihatan mirip penyebutnya yaitu:
$\begin{align}
& \dfrac{1}{10^{-2017}+1}+\dfrac{1}{10^{2017}+1} \\
&=\dfrac{10^{2017}+1}{(10^{-2017}+1)(10^{2017}+1)}+\dfrac{10^{-2017}+1}{(10^{-2017}+1)(10^{2017}+1)} \\
&=\dfrac{10^{2017}+1+10^{-2017}+1}{10^0+10^{2017}+10^{-2017}+1} \\
&=\dfrac{2+10^{2017}+10^{-2017}}{1+10^{2017}+10^{-2017}+1} \\
&=\dfrac{2+10^{2017}+10^{-2017}}{2+10^{2017}+10^{-2017}} \\
&=1
\end{align}$

$\begin{align}
& \dfrac{1}{10^{-2016}+1}+\dfrac{1}{10^{2016}+1} \\
&=\dfrac{10^{2016}+1}{(10^{-2016}+1)(10^{2016}+1)}+\dfrac{10^{-2016}+1}{(10^{-2016}+1)(10^{2016}+1)} \\
&=\dfrac{10^{2016}+1+10^{-2016}+1}{10^0+10^{2016}+10^{-2016}+1} \\
&=\dfrac{2+10^{2016}+10^{-2016}}{1+10^{2016}+10^{-2016}+1} \\
&=\dfrac{2+10^{2016}+10^{-2016}}{2+10^{2016}+10^{-2016}} \\
&=1
\end{align}$

$\begin{align}
& \dfrac{1}{10^{-2015}+1}+\dfrac{1}{10^{2015}+1} \\
&=\dfrac{10^{2015}+1}{(10^{-2015}+1)(10^{2015}+1)}+\dfrac{10^{-2015}+1}{(10^{-2015}+1)(10^{2015}+1)} \\
&=\dfrac{10^{2015}+1+10^{-2015}+1}{10^0+10^{2015}+10^{-2015}+1} \\
&=\dfrac{2+10^{2015}+10^{-2015}}{1+10^{2015}+10^{-2015}+1} \\
&=\dfrac{2+10^{2015}+10^{-2015}}{2+10^{2015}+10^{-2015}} \\
& =1
\end{align}$

Dari hasil diatas, jika kita jumlahkan dua pasangan pecahan yang penyebutnya "kelihatan hampir sama" maka kita peroleh hasilnya adalah $1$, dan soal diatas ada sebanyak $2017$ pasangan bilangan.

Pecahan $\dfrac{1}{10^{0}+1}$ tidak punya pasangan, tetapi nilainya dapat kita hitung yaitu $\dfrac{1}{10^{0}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}$. Hasil akhir dari soal diatas adalah $2017+\dfrac{1}{2}=2017,5$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 2017,5$

12. Soal UMB-PT 2014 Kode 573 (*Soal Lengkap)Solusi persamaan $5^{2x+1}=10^{2x-1}$ adalah...
$\begin{align}
(A)\ & {}^2\!\log 25 \\
(B)\ & {}^2\!\log 50 \\
(C)\ & {}^4\!\log 25 \\
(D)\ & {}^4\!\log 50 \\
(E)\ & {}^5\!\log 4
\end{align}$Alternatif Pembahasan: show

Bentuk persamaan kita coba manipulasi dengan sifat-sifat aljabar, seperti berikut ini;
$\begin{align}
5^{2x+1} &= 10^{2x-1} \\
5^{2x} \cdot 5^{1} &= 10^{2x} \cdot 10^{-1}\ (\times 10) \\
5^{2x} \cdot 50 &= 10^{2x} \\
50 & = \dfrac{10^{2x}}{5^{2x}} \\
50 & = \left( \dfrac{10}{5}\right)^{2x} \\
50 & = 2^{2x} \\
50 & = 4^{x} \\
\end{align}$
Dengan sedikit sentuhan dari logaritma yaitu $a^c=b \Leftrightarrow {}^a\!\log b=c$ maka dapat kita simpulkan $50 = 4^{x} \Leftrightarrow {}^4\!\log 50=x$

$\therefore$ Pilihan yang sesuai adalah $(D)\ {}^4\!\log 50 $

13. Soal UM UGM 2014 Kode 521 (*Soal Lengkap)Bentuk sederhana dari
$\dfrac{\left (x^{\frac{1}{3}}-x^{\frac{1}{6}} \right )\left (x^{\frac{1}{2}}+x \right )\left (x^{\frac{1}{2}}+x^{\frac{1}{3}}+x^{\frac{2}{3}} \right )}{\left (x^{\frac{4}{3}}-x \right )\left (x+x^{\frac{1}{3}}+x^{\frac{2}{3}} \right )}$ dengan $x \neq 0$ adalah...
$\begin{align}
(A)\ & x^{-\frac{1}{3}} \\
(B)\ & x^{\frac{1}{3}} \\
(C)\ & x^{\frac{2}{3}} \\
(D)\ & x^{-\frac{2}{3}} \\
(E)\ & x^{\frac{1}{2}}
\end{align}$Alternatif Pembahasan: show

Pangkat pecahannya coba kita samakan penyebutnya terlebih dahulu, bisar lebih cepat proses penjumlahannya;
$\begin{align}
& \dfrac{\left (x^{\frac{1}{3}}-x^{\frac{1}{6}} \right )\left (x^{\frac{1}{2}}+x \right )\left (x^{\frac{1}{2}}+x^{\frac{1}{3}}+x^{\frac{2}{3}} \right )}{\left (x^{\frac{4}{3}}-x \right )\left (x+x^{\frac{1}{3}}+x^{\frac{2}{3}} \right )} \\
& = \dfrac{\left (x^{\frac{2}{6}}-x^{\frac{1}{6}} \right )\left (x^{\frac{3}{6}}+x^{\frac{6}{6}} \right )\left (x^{\frac{3}{6}}+x^{\frac{2}{6}}+x^{\frac{4}{6}} \right )}{\left (x^{\frac{8}{6}}-x^{\frac{6}{6}} \right )\left (x^{\frac{6}{6}}+x^{\frac{2}{6}}+x^{\frac{4}{6}} \right )} \\
& = \dfrac{\left (x^{\frac{5}{6}}+x^{\frac{8}{6}}-x^{\frac{4}{6}}-x^{\frac{7}{6}} \right ) \left (x^{\frac{3}{6}}+x^{\frac{2}{6}}+x^{\frac{4}{6}} \right )}{\left (x^{\frac{14}{6}}+x^{\frac{10}{6}}+x^{\frac{12}{6}}-x^{\frac{12}{6}}-x^{\frac{8}{6}}-x^{\frac{10}{6}}\right )} \\
& = \dfrac{ x^{\frac{8}{6}}+x^{\frac{7}{6}}+x^{\frac{9}{6}}+x^{\frac{11}{6}}+x^{\frac{10}{6}}+x^{\frac{12}{6}}-x^{\frac{7}{6}}-x^{\frac{6}{6}}-x^{\frac{8}{6}}-x^{\frac{10}{6}}-x^{\frac{9}{6}}-x^{\frac{11}{6}} }{x^{\frac{14}{6}} -x^{\frac{8}{6}} } \\

& = \dfrac{x^{\frac{12}{6}}-x^{\frac{6}{6}}}{x^{\frac{14}{6}} -x^{\frac{8}{6}} } \\

& = \dfrac{x^{\frac{6}{6}} \left (x^{\frac{6}{6}}-1 \right )}{x^{\frac{8}{6}}\left (x^{\frac{6}{6}} -1 \right ) } \\

& = \dfrac{x^{\frac{6}{6}} }{x^{\frac{8}{6}}} \\
& = x^{\frac{6-8}{6}}=x^{-\frac{2}{3}}
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ x^{-\frac{2}{3}}$

14. Soal SBMPTN 2014 Kode 622 (*Soal Lengkap)Jika $4^{x}-4^{x-1}=6$ maka $(2x)^x$ sama dengan
$\begin{align}
(A)\ & 3 \\
(B)\ & 3\sqrt{3} \\
(C)\ & 9 \\
(D)\ & 9\sqrt{3} \\
(E)\ & 27
\end{align}$Alternatif Pembahasan: show

$\begin{align}
4^{x}-4^{x-1} & = 6 \\
4^{x}-4^{x} \cdot 4^{-1} & = 6\ (\times 4)\\
4 \cdot 4^{x}- 4^{x} & = 24 \\
4^{x} \left( 4 - 1 \right) & = 24 \\
4^{x} \left(3 \right) & = 24 \\
4^{x} & = 8 \\
2^{2x} & = 2^{3} \\
2x & = 3\ \Rightarrow x=\dfrac{3}{2}
\end{align}$

$\begin{align}
(2x)^{x} & = \left( 2 \cdot \dfrac{3}{2} \right)^{\dfrac{3}{2}} \\
& = \left( 3 \right)^{\dfrac{3}{2}} \\
& = 3 \cdot 3^{\dfrac{1}{2}} \\
& = 3 \sqrt{3}
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 3\sqrt{3}$

15. Soal SIMAK UI 2013 Kode 334 (*Soal Lengkap)Diketahui bahwa $3^{(y-x)}(x+y)=1$ dan $(x+y)^{(x-y)}=3$, nilai $x^{3y}=\cdots$
$\begin{align}
(1)\ & -\dfrac{1}{9} \\
(2)\ & \dfrac{1}{9} \\
(3)\ & 2 \\
(4)\ & 8
\end{align}$Alternatif Pembahasan: show

$\begin{align}
3^{(y-x)}(x+y) & = 1 \\
(x+y) & = \dfrac{1}{3^{(y-x)}} \\
(x+y) & = 3^{-(y-x)} \\
(x+y) & = 3^{(x-y)} \\
(x+y)^{(x-y)} & = 3 \\
3^{(x-y)^{(x-y)}} & = 3 \\
3^{(x-y)(x-y)} & = 3 \\
(x-y)^{2} & = 1\ \\
(x-y) & = \pm 1
\end{align}$

$\begin{align}
(x-y)=1 \rightarrow (x+y)^{(x-y)} & = 3 \\
(x+y)^{1} & = 3^{1} \\
(x+y) & = 3 \\
(x-y)=-1 \rightarrow (x+y)^{(x-y)} & = 3 \\
(x+y)^{-1} & = 3^{1} \\
(x+y) & = \dfrac{1}{3}
\end{align}$

$\begin{array}{c|c|cc}
x-y = 1 & \\
x+y = 3 & (+)\\
\hline
2x = 4 & \\
x = 2 & y=1\\
\hline
x^{3y} = 2^{3(1)} =8
\end{array} $

$\begin{array}{c|c|cc}
x-y = -1 & \\
x+y = \dfrac{1}{3} & (+) \\
\hline
2x = -\dfrac{2}{3} & \\
x = -\dfrac{1}{3} & y= \dfrac{2}{3}\\
\hline
x^{3y} = \dfrac{1}{3}^{3 \left( \frac{2}{3} \right)} =\dfrac{1}{9}
\end{array} $

$\therefore$ Pilihan yang sesuai adalah $(C)\ (2)\ \dfrac{1}{9} \text{dan}\ (4)\ 8$

16. Soal SIMAK UI 2013 Kode 332 (*Soal Lengkap)Jika $2^{(x+2)}+4^{(x+1)}=48$ nilai dari $\dfrac{1}{x+1} =\cdots$
$\begin{align}
(A)\ & { }^3\!\log 2 \\
(B)\ & \dfrac{1}{14} \\
(C)\ & { }^2\!\log 3 \\
(D)\ & { }^2\!\log 6 \\
(E)\ & 3 \\
\end{align}$Alternatif Pembahasan: show

$\begin{align}
2^{(x+2)}+4^{(x+1)} & = 48 \\
2^{x} \cdot 2^{2}+4^x \cdot \cdot 4^{1} & = 48 \\
2^{x} +4^x & = 12 \\
2^{x} +2^(2x) & = 12 \\
2^{x} \left(1+2^{x} \right) & = 12 \\
2^{x} \left(2^{x}+1 \right) & = 3(4) \\
2^{x} & = 3 \\
x & = { }^2\!\log 3
\end{align}$

Jika cara di atas kurang paham, coba alternatif berikut:
Saat $2^{x} +2^(2x)= 12$ kita misalkan $a=2^{x}$
$\begin{align}
2^{x} +2^(2x) & = 12 \\
a +a^(2) & = 12 \\
a^(2)+a-12 & = 0 \\
(a+4)(a-3) & = 0 \\
a & = -4\ \\
2^{x} & = -4\ (TM) \\
a & = 3 \\
2^{x} & = 3 \\
x & = { }^2\!\log 3
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(C)\ { }^2\!\log 3$

17. Soal UM UGM 2013 Kode 251 (*Soal Lengkap)Nilai $1-x$ yang memenuhi persamaan $\sqrt{8^{3-x}}=4 \cdot 2^{1-2x}$ adalah...
$\begin{align}
(A)\ & -4 \\
(B)\ & -3 \\
(C)\ & -2 \\
(D)\ & 3 \\
(E)\ & 4 \\
\end{align}$Alternatif Pembahasan: show

$\begin{align}
\sqrt{8^{3-x}} &= 4 \cdot 2^{1-2x} \\
8^{\dfrac{3-x}{2}} &= 2^{2} \cdot 2^{1-2x} \\
2^{ \dfrac{3(3-x)}{2}} &= 3-2x \\
9-3x &= 6-4x \\
4x-3x &= 6-9 \\
x &= -3 \\
1- x &= 1-(-3) =4
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 4$

18. Soal SBMPTN 2013 Kode 327 (*Soal Lengkap)Jika $8^{m}=27$, maka $2^{m+2}+4^{m}=\cdots$
$\begin{align}
(A)\ & 12 \\
(B)\ & 15 \\
(C)\ & 18 \\
(D)\ & 21 \\
(E)\ & 24
\end{align}$Alternatif Pembahasan: show

$\begin{align}
8^{m} & = 27 \\
m & = { }^8\!\log 27 \\
m & = { }^{2^{3}}\!\log 3^{3} \\
m & = \dfrac{3}{3} \cdot { }^2 \!\log 3 \\
m & = { }^2\!\log 3 \\
2^{m+2}+4^{m} & = 2^{m} \cdot 2^{2} + 2^{2m} \\
& = 2^{{ }^2\!\log 3} \cdot 4 + 2^{2 \cdot { }^2\!\log 3} \\
& = 3 \cdot 4 + 2^{ { }^2\!\log 3^{2}} \\
& = 12 + 9=21
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ 21$

19. Soal SBMPTN 2013 Kode 125 (*Soal Lengkap)Jika $9^{m-1}+9^{m+1}=82$, maka $4^{m+1}=\cdots$
$\begin{align}
(A)\ & \dfrac{1}{16} \\
(B)\ & \dfrac{1}{4} \\
(C)\ & 4 \\
(D)\ & 16 \\
(E)\ & 64
\end{align}$Alternatif Pembahasan: show

$\begin{align}
9^{m-1}+9^{m+1} & = 82 \\
9^{m} \cdot 9^{-1}+9^{m} \cdot 9^{ 1} & = 82 \\
9^{m} \left( 9^{-1}+ 9 \right) & = 82 \\
9^{m} \left( \dfrac{82}{9} \right) & = 82 \\
9^{m} & = 82 \cdot \left( \dfrac{9}{82} \right) \\
9^{m} & = 9 \\
m & = 1 \\
4^{m+1} & = 4^{1+1}=16
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ 16$

20. Soal SIMAK UI 2012 Kode 222 (*Soal lengkap)$\dfrac{5^{4022}-5^{4018}}{5^{4020}-5^{4016}}=\cdots$
$\begin{align}
(A)\ & 1 \\
(B)\ & 3 \\
(C)\ & \dfrac{25}{4} \\
(D)\ & \dfrac{25}{2} \\
(E)\ & 25
\end{align}$Alternatif Pembahasan: show

$\begin{align}
\dfrac{5^{4022}-5^{4018}}{5^{4020}-5^{4016}} &= \dfrac{5^{4018} \left( 5^{4} -1 \right)}{5^{4016} \left( 5^{4} -1 \right)} \\
&= \dfrac{5^{4018} }{5^{4016} } \\
&= 5^{4018-4016} \\
&= 5^{2}=25
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(E)\ 25$

21. Soal SIMAK UI 2012 Kode 221 (*Soal lengkap)Hasil perkalian dari nilai $x$ yang memenuhi $\dfrac{x^{2}}{10.000}=\dfrac{10.000}{x^{2 \left({}^{10}\!\log x \right)-8}}$ adalah...
$\begin{align}
(A)\ & 10^{2} \\
(B)\ & 10^{3} \\
(C)\ & 10^{4} \\
(D)\ & 10^{5} \\
(E)\ & 10^{7}
\end{align}$Alternatif Pembahasan: show

$\begin{align}
\dfrac{x^{2}}{10.000} &= \dfrac{10.000}{x^{2 \left( {}^{10}\!\log x \right)} -8} \\
x^{2} \cdot x^{2 \left({}^{10}\!\log x \right)-8} &= 10^{8} \\
x^{2 \left({}^{10}\!\log x \right)-6} &= 10^{8} \\
{}^x\!\log 10^{8} &= {}^{10}\!\log x -6 \\
8 {}^x\!\log 10 &= 2 {}^{10}\!\log x -6
\end{align}$
Misal: $m={}^x\!\log 10$ maka $\dfrac{1}{m}={}^{10}\!\log x$
$\begin{align}
8m &= \dfrac{2}{m}-6 \\
8m^{2} &= 2-6m \\
4m^{2}+3m-1 &= 0 \\
(4m-1)( m+1) &= 0 \\
m = -1\ \text{atau}\ m &= \dfrac{1}{4}
\end{align}$

$\begin{align}
m &= {}^x\!\log 10\ \Rightarrow \dfrac{1}{4} = {}^x\!\log 10 \\
x^{\dfrac{1}{4}} &= 10 \Rightarrow x = 10^{4}
\end{align}$

$\begin{align}
m &= {}^x\!\log 10 \Rightarrow -1 = {}^x\!\log 10 \\
x^{-1} &= 10 \Rightarrow x = 10^{-1}
\end{align}$

Hasil perkalian nilai $x$ adalah $10^{4} \cdot 10^{-1}=10^{3}$

$\therefore$ Pilihan yang sesuai adalah $(B)\ 10^{3}$

22. Soal SNMPTN 2012 Kode 121 (*Soal Lengkap)Jika $a$ dan $b$ adalah bilangan bulat positif yang memenuhi $a^{b}=2^{20}-2^{19}$, maka nilai $a+b$ adalah...
$\begin{align}
(A)\ & 3 \\
(B)\ & 7 \\
(C)\ & 19 \\
(D)\ & 21 \\
(E)\ & 23
\end{align}$Alternatif Pembahasan: show

Bentuk persamaan kita coba manipulasi dengan sifat-sifat aljabar, seperti berikut ini;
$\begin{align}
a^{b} &= 2^{20}-2^{19} \\
&= 2^{19} \left( 2-1 \right) \\
&= 2^{19} \\
a &= 2 \\
b &= 19 \\
a+b &= 19+2=21
\end{align}$

$\therefore$ Pilihan yang sesuai adalah $(D)\ 21 $

Jika engkau tidak sanggup menahan lelahnya belajar, Maka engkau harus menanggung pahitnya kebodohan ___pythagoras

lembar jawaban penilaian harian matematika,

lembar jawaban penilaian akhir semester matematika,

presentasi hasil diskusi matematika atau

pembahasan quiz matematika di kelas.

CMIIWJADIKAN HARI INI LUAR BIASA! - WITH GOD ALL THINGS ARE POSSIBLE

atematika Dasar yang akan kita diskusikan berikut adalah tentang Eksponen atau Bilangan Be Bank Soal Matematika Dasar Eksponen (*Soal dan Pembahasan)

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