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

fertilizeki 2021-12-31 Answered
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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Expert Answer

yotaniwc
Answered 2022-01-01 Author has 4798 answers
... 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 \(\displaystyle{x}={\frac{{{6}}}{{{4}}}}={\frac{{{3}}}{{{2}}}}\).
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Nadine Salcido
Answered 2022-01-02 Author has 1327 answers
\(\displaystyle{2}^{{{4}{x}+{1}}}={128}\Leftrightarrow\)
\(\displaystyle{{{\log}_{{22}}^{{{4}{x}+{1}}}}=}{{\log}_{{2128}}\Leftrightarrow}\)
\(\displaystyle{4}{x}+{1}={7}\Leftrightarrow\)
\(\displaystyle{4}{x}={6}\Leftrightarrow\)
\(\displaystyle{x}={\frac{{{6}}}{{{4}}}}\)
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Vasquez
Answered 2022-01-09 Author has 8850 answers

\(2^{4x+1}=128\)
\(\log2^{2x+1}=\log128\)
\((4x+1)\times\log2=\log128\text{ -from properties of logs}\)
\(x=\frac{1}{4}(\frac{\log128}{\log(2)}-1)=\frac{3}{2}\)
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

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