$16^{18}+16^{18}+16^{18}+16^{18}+16^{18}=4^x$ without a calculator

Question

How to solve the following equation without using calculator

$$16^{18}+16^{18}+16^{18}+16^{18}+16^{18}=4^x$$

Answer 1

$$16^{18}+16^{18}+16^{18}+16^{18}+16^{18}=5\cdot 16^{18}=5\cdot 4^{36}=4^x$$

The solution for $\,x\,$ is going to be a little ugly because of that $\,5\,$ there. If instead of $\,5\,$ summands there were only $\,4\,$ then things would be nicer...anyway:

$$5\cdot 4^{36}=4^x\Longrightarrow x=\frac{\log 5+36\log 4}{\log 4}=\frac{\log 5}{\log 4}+36$$

Answer 2

Left hand side is $$5\cdot 16^{18} = 4^{\log_45}\cdot 16^{18} $$ So, because $16=4^2$, we get $x=\log_45 + 2\cdot 18$.

Answer 3

$5\cdot 16^{18}=4^x\implies 5\cdot (2^4)^{18}=(2^2)^x \implies5\cdot 2^{72}=2^{2x}$

So, $5=2^{2x-72}$

Taking logarithm with base $2,\log_25=2x-72\implies x=36+\frac{\log_25}2$

Answer 4

$16^{18}+16^{18}+16^{18}+16^{18}+16^{18}$

$=5\times 16^{18}$

$= 5\times {4^2}^{18} $

$= 5\times 4^{2\times 18} = 5\times 4^{36} = 4^x$

$ 5\times 4^{36} = 4^{\log_4{5}} \times 4^{36} $ $= 4^{\log_4{5} +36}$

$\rightarrow x= \log_4{5} +36$