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

How to solve the following equation without using calculator

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

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Start with law of exponents $16^{18}=4^{?}$ – Maesumi Oct 8 '12 at 13:05

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

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:

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

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Nice using of summation, Don. +) – Babak S. Aug 11 '13 at 14:43

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

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

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

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

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For an approximation to $\log_2 5$ you can use that $2^7=128 \approx 5^3=125$ and taking logs to base 2 this gives $7 \approx 3 \log_2 5$ – Mark Bennet Oct 8 '12 at 13:12
@MarkBennet: Yes. Well done. – Ross Millikan Oct 8 '12 at 13:15
@MarkBennet, thanks for your input. SO, $x≈36+\frac{7}{6}=37\frac 1 6$ – lab bhattacharjee Oct 8 '12 at 13:19

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

$=5\times 16^{18}$

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

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

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

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

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