fertilizeki

2021-12-31

How to find the value of an unknown exponent?
E.g. I have the question:
${2}^{4x+1}=128$
I solved this by knowing that $128={2}^{7}$ and therefore x must equal $1.5$.
However, is there a way of solving this without knowing that $128={2}^{7}$?

yotaniwc

Expert

... and without logarithms or knowing any powers of 2 other than the most trivial one ...
${2}^{4x+1}=128$
${2}^{4x+1-1}={2}^{4x}=\frac{128}{2}=64$
${2}^{4x-1}=32$
${2}^{4x-2}=16$
${2}^{4x-3}=8$
${2}^{4x-4}=4$
${2}^{4x-5}=2$
${2}^{4x-6}=1={2}^{0}$
so $4x-6=0$ and $x=\frac{6}{4}=\frac{3}{2}$.

Expert

${2}^{4x+1}=128⇔$
${\mathrm{log}}_{22}^{4x+1}={\mathrm{log}}_{2128}⇔$
$4x+1=7⇔$
$4x=6⇔$
$x=\frac{6}{4}$

Vasquez

Expert

${2}^{4x+1}=128$
$\mathrm{log}{2}^{2x+1}=\mathrm{log}128$

$x=\frac{1}{4}\left(\frac{\mathrm{log}128}{\mathrm{log}\left(2\right)}-1\right)=\frac{3}{2}$
note that you can use any logarithm, log base 10 or '$\mathrm{ln}$' - or any other 'base' of logarithms you might have (with log10 and loge being the commonly found ones on calculators, spreadsheets etc ) you have to use your chosen type of log consistently of course

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