Contoh Soal Fungsi Ekponensial dan Logaritma

1. Jika y=3^{x} , nyatakan bentuk berikut ini dalam bentuk y

\begin{matrix} a. & 3^{x+1} & - & 3^{x-1}\\ b. & 3^{2x+1} & - &3^{x-2} \\ c. & 2\left ( 3^{1-x} \right )\\ d. & 3^{x} & + & 3^{x+1}&+&3^{x+2}\\ e. & 9^{x} & - & 27^{\frac{1}{3}x+\frac{2}{3}}\\ f. & 3^{x} & - & 9^{\frac{1}{2}x+1}&+&27^{\frac{1}{3}\left ( x+2 \right )} \end{matrix}

Jawab:

 a. 3^{x+1}-3^{x-1}=3^{x}.3-3^{x}.\frac{1}{3}=\left ( 3-\frac{1}{3} \right )3^{x}=\frac{8}{3}3^{x}=\frac{8}{3}y.

b. 3^{2x+1}-3^{x-2}=3^{2x}.3-3^{x}.\frac{1}{9}=3.\left ( 3^{x} \right )^{2}-\frac{3^{x}}{9}=3y^{2}-\frac{1}{9}y.

c. 2\left ( 3^{1-x} \right )=2.\frac{3}{3^{x}}=\frac{6}{3^{x}}=\frac{6}{y}

d. 3^{x}+3^{x+1}+3^{x+2}=3^{x}+3.3^{x}+9.3^{x}=\left ( 1+3+9 \right ).3^{x}=13y

e. 9^{x}-27^{\frac{1}{3}x+\frac{2}{3}}=3^{2x}-3^{3\left ( \frac{1}{3}x \right )+\frac{2}{3}}=\left ( 3^{x} \right )^{2}-3^{x+2}=y^{2}-9y

f. 3^{x}-9^{\frac{1}{2}x+1}+27^{\frac{1}{3}\left ( x+2 \right )}=3^{x}-3^{2\left ( \frac{1}{2}x+1 \right )}+3^{3.\frac{1}{3}\left ( x+2 \right )}=3^{x}-3^{x+2}+3^{x+2}=3^{x}=y.

2. Jika p=2^{2x} , nyatakan bentuk berikut dalam bentuk p

\begin{matrix} a. & 2^{2x} & + & 3\left ( 2^{-2x} \right )\\ b. & 4^{x} & - & 3\left ( 16^{x} \right )\\ c. & 2^{2x+1} & + & 4^{x-1}\\ d. & 8^{\frac{2}{3}x+\frac{1}{3}} & + & 4^{x+1}\\ e. & 2^{2x} & + & 2^{2x+1}&+&4^{x+2}\\ f. & 2^{2x-1} & - & 4^{2x+1}&+&16^{x-1} \end{matrix}

Jawab:

a. 2^{2x}+3\left ( 2^{-2x} \right )=2^{2x}+\frac{3}{2^{2x}}=p+\frac{3}{p}=\frac{p^{2}+3}{p}.

b. 4^{x}-3\left ( 16^{x} \right )=2^{2x}-3\left ( 2^{4x} \right )=2^{2x}-3.\left ( 2^{2x} \right )^{2}=p-3p^{2}.

c. 2^{2x+1}+4^{x-1}=2.2^{2x}+\frac{4^{x}}{4}=2p+\frac{p}{4}=\frac{9}{4}p.

d, e, dan f  sebagai latihan.

3. Sederhanakanlah

\begin{matrix} a. & 3^{x+4} &.& 5^{x+1}&.&15^{2x-1} \\ b. & 6^{3x+1} &.&8^{x-1} &. &24^{3x-1} \\ c. & 5^{x+7} &.&25^{2x-1}&. &125^{2-x} \\ d. & 2^{x-1} &.&4^{3x-2}&. &32^{2x+1} \\ e. & \left ( 6^{x}.12^{2x+2} \right ) &:& \left ( 27^{x}.32^{3x} \right ) \\ f.&20^{x+3} &.& 15^{2x+5} & .&6^{2x-1} \end{matrix}

Jawab:

a. 3^{x+4}.5^{x+1}.15^{2x-1}=3^{x}.3^{4}.5^{x}.5.(3.5)^{2x}.\frac{1}{15}=27.15^{3x}=3^{3}.\left ( 15^{x} \right )^{3}=\left ( 3.15^{x} \right )^{3}

b. 6^{3x+1}.8^{x-1}.24^{3x-1}=6^{3x}.6.2^{3x-3}.\frac{24^{3x}}{24}=\frac{1}{8}.\frac{6}{24}.\left ( 2.3 \right )^{3x}.2^{3x}.\left ( 2^{3}.3 \right )^{3x}=\frac{1}{32}.2^{3x+3x+9x}.3^{3x+3x}=\frac{1}{32}.2^{15x}.3^{6x}=2^{15x-5}.3^{6x}

c. 5^{x+7}.25^{2x-1}.125^{2-x}=5^{x+7}.5^{4x-2}.5^{6-3x}=5^{x+4x-3x+7-2+6}=5^{2x+11}

d, e, dan f sebagai latihan

4. Jika diberikan  f(x)=8^{x}, maka nilai dari

\begin{matrix} a. & f\left ( \frac{a}{3} \right )\\ b. & f\left ( \frac{1}{\sqrt{3}} \right ) \end{matrix}

Jawab:

Dari soal diketahui  f\left ( x \right )=8^{x}=\left ( 2^{3} \right )^{x}=\left ( 2^{x} \right )^{3}

a. f\left ( \frac{a}{3} \right )=\left ( 2^{\frac{a}{3}} \right )^{3}=2^{a}

b. f\left ( \frac{1}{\sqrt{3}} \right )=\left ( 2^{\frac{1}{\sqrt{3}}} \right )^{3}=\left ( 2^{\frac{1}{\sqrt{3}}} \right )^{\sqrt{3}.\sqrt{3}}=2^{\sqrt{3}}

5. Jika diketahui

\left\{\begin{matrix} f\left ( x \right ) & = & \frac{1}{2}\left ( a^{x} +a^{-x}\right ) \\ g\left ( x \right ) & = & \frac{1}{2}\left ( a^{x}-a^{-x} \right ) \end{matrix}\right.

Tunjukkan bahwa   f^{2}\left ( x \right )-g^{2}\left ( x \right )=1

Jawab:

\left ( f\left ( x \right ) \right )^{2}=\left ( \frac{1}{2} \left ( a^{x}+a^{-x} \right )\right )^{2}=\frac{1}{4}\left ( a^{2x}+2+a^{-2x} \right ).

\left ( g\left ( x \right ) \right )^{2}=\left ( \frac{1}{2} \left ( a^{x}-a^{-x} \right )\right )^{2}=\frac{1}{4}\left ( a^{2x}-2+a^{-2x} \right ).

sehingga

\left ( f\left ( x \right ) \right )^{2}-\left ( g\left ( x \right ) \right )^{2}=\frac{1}{4}\left ( a^{2x}+2+a^{-2x} \right )-\frac{1}{4}\left ( a^{2x}-2+a^{-2x} \right )=\frac{2}{4}+\frac{2}{4}=\frac{4}{4}=1

terbukti

6. Diketahui f\left ( x \right )=a^{x}+a^{-x}  dan g\left ( x \right )=a^{x}-a^{-x}. Tentukanlah

\begin{matrix} a. & \left ( f\left ( x \right ) \right )^{2} &-&\left ( g\left ( x \right ) \right )^{2} \\ b. & g\left ( 2x \right ) &-&f\left ( x \right ).g\left ( x \right ) \end{matrix}

Jawab:

a. \left ( f\left ( x \right ) \right )^{2}-\left ( g\left ( x \right ) \right )^{2}=\left ( a^{2x}+2+a^{-2x} \right )-\left ( a^{2x}-2+a^{-2x} \right )=2+2=4.

b. g\left ( 2x \right )-f\left ( x \right ).g\left ( x \right )=a^{2x}-a^{-2x}-\left ( a^{x}+a^{-x} \right )\left ( a^{x} -a^{-x}\right )=a^{2x}-a^{-2x}-\left ( a^{2x}-a^{-2x} \right )=0.

7. Diberikan f\left ( x \right )=2^{x}. Tunjukkan bahwa

\frac{f\left ( x+h \right )-f\left ( x \right )}{h}=2^{x}\left ( \frac{2^{h}-1}{h} \right )

Jawab:

\frac{f\left ( x+h \right )-f\left ( x \right )}{h}=\frac{2^{x+h}-2^{x}}{h}=\frac{2^{x}.2^{h}-2^{x}}{h}=2^{x}\left ( \frac{2^{h}-1}{h} \right ).

8. Diketahui \Phi \left ( t \right )=a^{t}+1 . Tunjukkan bahwa

\frac{1}{\Phi \left ( t \right )}+\frac{1}{\Phi \left ( -t \right )}=1

Jawab:

\frac{1}{\Phi \left ( t \right )}+\frac{1}{\Phi \left ( -t \right )}=\frac{1}{a^{t}+1}+\frac{1}{a^{-t}+1}=\frac{1}{1+a^{t}}+\frac{1}{\frac{1}{a^{t}}+1}=\frac{1}{1+a^{t}}+\frac{a^{t}}{1+a^{t}}=\frac{1+a^{t}}{1+a^{t}}=1.

9. Jika a^{2x}=\sqrt{2}-1, maka nilai dari

\begin{matrix} a. & a^{x} &+&a^{-x} \\ b. & \frac{a^{3x}+a^{-3x}}{a^{x}+a^{-x}} \end{matrix}

Jawab:

a.  perhatikan bahwa

\left ( a^{x}+a^{-x} \right )^{2}=a^{2x}+2+a^{-2x}\: \Rightarrow \: a^{x}+a^{-x}=\sqrt{a^{2x}+2+\frac{1}{a^{2x}}}=\sqrt{\sqrt{2}-1+2+\frac{1}{\sqrt{2}-1}}=\sqrt{\sqrt{2}-1+2+\sqrt{2}+1}=\sqrt{2\sqrt{2}+2}.

b. karena\: \: \frac{a^{3x}+a^{-3x}}{a^{x}+a^{-x}}=a^{2x}-1+a^{-2x}=a^{2x}-1+\frac{1}{a^{2x}} , sehingga

a^{2x}-1+\frac{1}{a^{2x}}=\sqrt{2}-1-1+\frac{1}{\sqrt{2}-1}=\sqrt{2}-1-1+\sqrt{2}+1=2\sqrt{2}-1.

10. Diketahui f\left ( x \right )=1-\frac{1}{x}\: ,\: f^{2}\left ( x \right )=f\left ( f\left ( x \right ) \right )\: ,dan\: \: f^{n}\left ( x \right )=f\left ( f^{n-1}\left ( x \right ) \right )  untuk n=3,4,5,....

Tentukanlah nilai dari f^{2014}\left ( \frac{1}{2} \right )

Jawab:

Perhatikan bahwa f\left ( \frac{1}{2} \right )=1-\frac{1}{\frac{1}{2}}=1-2=-1.

Selanjutnya

\begin{matrix} f\left ( \frac{1}{2} \right ) & = &1-\frac{1}{\frac{1}{2}}= 1-2=-1\\ f^{2}\left ( \frac{1}{2} \right ) & = &f\left ( f\left ( \frac{1}{2} \right ) \right )=f\left ( -1 \right )=2 \\ f^{3}\left ( \frac{1}{2} \right ) & = & f\left ( f^{2}\left ( \frac{1}{2} \right ) \right )=f\left ( 2 \right )=\frac{1}{2}\\ f^{4}\left ( \frac{1}{2} \right ) & = & f\left ( f^{3}\left ( \frac{1}{2} \right ) \right )=f\left ( \frac{1}{2} \right )=-1\\ f^{5}\left ( \frac{1}{2} \right ) & = & f\left ( f^{4}\left ( \frac{1}{2} \right ) \right )=f\left ( -1 \right )=2\\ \vdots &=&\vdots&\cdots \\ \vdots &=&\vdots&\cdots \\ f^{2014}\left ( \frac{1}{2} \right )&=&\cdots =f^{4}\left ( \frac{1}{2} \right )=f\left ( \frac{1}{2} \right )=-1 \end{matrix}

Sumber Referensi

  1. Sukino. 2013. Matematika Kelompok Peminatan Matematika dan Ilmu Alam untuk SMA/MA Kelas x. Jakarta: Erlangga.

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