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May 26, 2018 Bank Soal, Eksponen, Matematika SMA, Rumus Matematika, SIMAK UI, UM UGM

Matematika Dasar Eksponen/ Bilangan Berpangkat (*Soal Dari Aneka Macam Sumber)

Kesulitan 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. SNMPTN 2010 Kode 336Jika $n$ 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:

Hint

$25^{0.25} \times 25^{0.25}\times \cdots \times 25^{0.25}\times 25^{0.25}=125$
$5^{0.5} \times 25^{0.5}\times \cdots \times 5^{0.5}\times 25^{0.5}=5^{3}$
$5^{0.5} \times 5^{0.5}\times \cdots \times 5^{0.5}\times 5^{0.5}=5^{3}$
$\left(5^{0.5}\right)^{n}=5^{3}$
$5^{\dfrac{1}{2}n}=5^{3}$
$\dfrac{1}{2}n=3$
$n=6$

$(n-3)(n+2)=(6-3)(6+2)=24$
$\therefore$ Pilihan yang sesuai yaitu $(A)\ 24$

2. SIMAK UI 2009 Kode 951 [Soal SIMAK UI lengkap]Diketahui $x_{0}$ dan $y_{0}$ yaitu 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$\cdots$
$(A)\ -1$
$(B)\ 0$
$(C)\ 1$
$(D)\ 2$
$(E)\ 4$Alternatif Pembahasan:

Hint

$\begin{cases}2^{x+1}-3^{y}=7 \\ -\left(2^{x-1} \right)-3^{y+1}=-5\end{cases}$
$\begin{cases}2^{x} \cdot 2^{1}-3^{y}=7 \\ 2^{x-1}+3^{y+1}=5\end{cases}$
$\begin{cases}2^{x} \cdot 2^{1}-3^{y}=7 \\ 2^{x} \cdot 2^{-1}+3^{y} \cdot 3^{1}=5\end{cases}$

Dengan memisalkan $m=2^{x}$ dan $n=3^{y}$, sistem persamaan sanggup kita ubah menjadi;
$\begin{cases}2m-n=7 \\ \dfrac{1}{2}m+3n=5\end{cases}$
$\begin{cases}2m-n=7 \\ 2m+12n=20\end{cases}$

Dari sistem persamaan diatas dengan proses eliminasi atau substitusi kita peroleh nilai $m=2$ dan $n=1$.
Kita kembali kepada pemisalan, $m=2^{x}$ $\Rightarrow$ $4=2^{x}$ $\Rightarrow$ $x=2$ dan
$n=3^{y}$ $\Rightarrow$ $1=3^{y}$ $\Rightarrow$ $y=0$

Nilai $x_{0}+y_{0}=2+0=2$

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

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

Hint

\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 yaitu $(E)\ 2$

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

Hint

Untuk menuntaskan problem ini, kita butuh beberapa hukum dari logaritma, salah satunya yaitu jikalau $b > 0$, dan $a > 0 $, $a \neq 1$ maka $a^c=b \Leftrightarrow {}^a\!\log b=c$.
$a^c=b \Leftrightarrow {}^a\!\log b=c$

$x^{y}=xy \Leftrightarrow {}^x\!\log (xy)=y$
${}^x\!\log (xy)=y$
${}^x\!\log x+{}^x\!\log y=y$
$1+{}^x\!\log y=y$
${}^x\!\log y=y-1$ $\cdots (pers.1)$

$\dfrac{x}{y}=x^{5y} \Leftrightarrow {}^x\!\log (\dfrac{x}{y})=5y$
${}^x\!\log (\dfrac{x}{y})=5y$
${}^x\!\log x-{}^x\!\log y=5y$
$1-{}^x\!\log y=5y$
${}^x\!\log y=1-5y$ $\cdots (pers.2)$

Dengan mensubstitusi $(pers.1)$ dan $(pers.2)$ maka kita peroleh $y-1=1-5y$ $\Leftrightarrow$ $y=\dfrac{1}{3}$
$xy=x^{y}$
$x \cdot \dfrac{1}{3} =x^{\dfrac{1}{3}}$
$x=3x^{\dfrac{1}{3}}$
$x^{3}=27x$
$x^{2}=27$

$x^{2}+3y=27+3(\dfrac{1}{3})=28$

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

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

Hint

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

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

6. SIMAK UI 2013 Kode 437 [Soal SIMAK UI 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 bundar faktual dan $w$ bilangan bundar nonnegative dengan $a \lt b \lt c$. Nilai $(2w)+(ax)+(by)+(cz)=\ldots$
$(A)\ 0$
$(B)\ 3$
$(C)\ 11$
$(D)\ 75$
$(E)\ 611$Alternatif Pembahasan:

Hint

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

Nilai $(2w)+(ax)+(by)+(cz)=\ldots$
$=(2 \cdot 0)+(3 \cdot 1)+(11 \cdot 1)+(61 \cdot 1)$
$= 0+3+11+61$
$=75$ $\D$

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

Hint

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

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

8. SIMAK UI 2014 Kode 511 [Soal SIMAK UI lengkap]Dalam basis 10, bilangan bundar faktual $p$ mempunyai $3$ digit, bilangan bundar faktual $q$ mempunyai $p$ digit, bilangan bundar faktual $r$ mempunyai $q$ digit. Nilai untuk terkecil untuk $r$ adalah$cdots$
$(A)\ 10^{10^{100}}$
$(B)\ 10^{10^{100}-1}$
$(C)\ 10^{10^{99}}$
$(D)\ 10^{10^{99}-1}$
$(E)\ 10^{99^{99}}$Alternatif Pembahasan:

Hint

Topik ini tolong-menolong tidak murni perihal eksponen, tetapi alasannya yaitu pilihannya bilangan berpangkat para siswa melihat ini perihal bilangan berpangkat. Ada sedikit logika atau teori bilangan didalamnya.

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

Bilangan $p$ terdiri dari $3$ digit, supaya mendapat $p$ 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 mendapat $q$ 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 mendapat $r$ bilangan terkecil maka angka pertama dipilih angka $1$ dan sisanya dipilih angka nol sehingga $r = 10^{10^{99}-1}$

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

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

Hint

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

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

10. SIMAK UI 2015 Kode 563 [Soal SIMAK UI lengkap]$\dfrac{2015^{2}(2014^{2}-2013)}{(2014^{2}-1)(2014^{3}+1)}\times \dfrac{2013^{2}(2014^{2}+2015)}{(2014^{3}-1)}=\cdots$
$(A)\ 2013 \times 2015$
$(B)\ 2015$
$(C)\ 2014$
$(D)\ 2013$
$(E)\ 1$Alternatif Pembahasan:

Hint

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

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

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

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