138th Problem 2017

$4^{x+1}$ =8

or, $2^{2x+2}$ = $2^3$

or,2x+2 =3

or, x = $\frac {1}{2}$

Solution 1:

$4^{x+1}=8$

$2^{2(x+1)}=2^3$

$2(x+1)=3$

$x+1=\dfrac{3}{2}$

$x=\dfrac{3}{2}-1$

$x=\dfrac{3}{2}-\dfrac{2}{2}$

$x=\dfrac{1}{2}$

$x=0.5$ $\large\color{#D61F06}\boxed{answer}$

Solution 2:

taking the natural logarithm of both sides

$4^{x+1}=8$

$\ln~4^{x+1}=\ln~8$

$(x+1)\ln~4=\ln~8$

$x+1=\dfrac{\ln~8}{\ln~2}$

By the use of a calculator, we get

$x+1=1.5$

Finally,

$x=0.5$ $\large\color{#D61F06}\boxed{answer}$

$\displaystyle 4^{x \ + \ 1} = 8$

$\displaystyle (2^2)^{x \ + \ 1} = 2^3$

$\displaystyle 2^{2(x \ + \ 1)} = 2^3$

$\displaystyle 2(x + 1) = 3$

$\displaystyle 2x = 1$

$\displaystyle x = \frac 12$