# How to find the value of an unknown exponent? E.g. I

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

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yotaniwc
... and without logarithms or knowing any powers of 2 other than the most trivial one ...
$$\displaystyle{2}^{{{4}{x}+{1}}}={128}$$
$$\displaystyle{2}^{{{4}{x}+{1}-{1}}}={2}^{{{4}{x}}}={\frac{{{128}}}{{{2}}}}={64}$$
$$\displaystyle{2}^{{{4}{x}-{1}}}={32}$$
$$\displaystyle{2}^{{{4}{x}-{2}}}={16}$$
$$\displaystyle{2}^{{{4}{x}-{3}}}={8}$$
$$\displaystyle{2}^{{{4}{x}-{4}}}={4}$$
$$\displaystyle{2}^{{{4}{x}-{5}}}={2}$$
$$\displaystyle{2}^{{{4}{x}-{6}}}={1}={2}^{{0}}$$
so $$\displaystyle{4}{x}−{6}={0}$$ and $$\displaystyle{x}={\frac{{{6}}}{{{4}}}}={\frac{{{3}}}{{{2}}}}$$.
###### Not exactly what you’re looking for?
$$\displaystyle{2}^{{{4}{x}+{1}}}={128}\Leftrightarrow$$
$$\displaystyle{{{\log}_{{22}}^{{{4}{x}+{1}}}}=}{{\log}_{{2128}}\Leftrightarrow}$$
$$\displaystyle{4}{x}+{1}={7}\Leftrightarrow$$
$$\displaystyle{4}{x}={6}\Leftrightarrow$$
$$\displaystyle{x}={\frac{{{6}}}{{{4}}}}$$
Vasquez

$$2^{4x+1}=128$$
$$\log2^{2x+1}=\log128$$
$$(4x+1)\times\log2=\log128\text{ -from properties of logs}$$
$$x=\frac{1}{4}(\frac{\log128}{\log(2)}-1)=\frac{3}{2}$$
note that you can use any logarithm, log base 10 or '$$\ln$$' - 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