How do you solve $5 (4^{36}) = 4^{x}$ $5(4^36)=4^x$ ?

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Answer 1 · 2016-05-07T14:09:42

$x = 37.161$ $x=37.161$

$5 (4^{36}) = 4^{x}$ $5(4^36)=4^x$ means

$x = {log}_{4} (5 (4^{36})$ $x=log_4(5(4^36)$ or

$x = {log}_{4} (5) + {log}_{4} (4^{36})$ $x=log_4(5)+log_4(4^36)$ or

$x = {log}_{4} (5) + 36 {log}_{4} (4)$ $x=log_4(5)+36log_4(4)$ or

$x = {log}_{4} 5 + 36$ $x=log_4 5+36$ or

$x = \frac{log 5}{log 4} + 36 = 1.161 + 36 = 37.161$ $x=log5/log4+36=1.161+36=37.161$

Answer 2 · 2016-05-09T07:57:33

Use laws of indices first, then logs.
$37.61 = x$ $37.61 = x$

There are powers of 4 on both sides of the equation.

$5 (4^{36}) = 4^{x} divide by 4^{36}$ $5(4^36) = 4^x " divide by" 4^36$

$5 = \frac{4^{x}}{4^{36}} subtract indices$ $5 = (4^x)/4^36" subtract indices"$

$5 = 4^{x - 36}$ $5 = 4^(x - 36)$

$log 5 = (x - 36) log 4$ $log 5 = (x - 36)log4$

$\frac{log 5}{log 4} = x - 36$ $log5/log4 = x - 36$

$1.6096 = x - 36$ $1.6096 = x - 36$

$37.61 = x$ $37.61 = x$

How do you solve 5(4^36)=4^x5(436)=4x5(4^36)=4^x?