# Convert $2^{25}$ to its equivalent in base 10 (i.e. $10^?$) [duplicate]

Is there a straightforward way to convert from one number to the power of something to its equivalent as another number to the power of something?

For example, how do you convert $2^{25}$ into $10^{something}$?

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## marked as duplicate by MJD, Asaf Karagila, rschwieb, Davide Giraudo, 5pm Feb 26 '13 at 20:44

logarithms and antilogarithms – hjpotter92 Feb 26 '13 at 18:07

Your question be be rewritten as follows: $10^x = 2^{25}$. Where $x$ is your "something".

The answer to this requires the use of logarithms. We can "take the log of both sides", and then use the definition of logarithms and few laws to get to an answer:

\begin{array}{ccc} 10^x &=& 2^{25} \\ \log_{10}(10^x) &=& \log_{10}(2^{25}) \\ x &=& 25\log_{10}(2) \\ x &=& 7.5257... \end{array}

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$$2^{25}=10^x$$ $$\log 2^{25}=\log10^x$$ $$25\log2=x\log10$$ $$x=\frac{25\log2}{\log10}$$

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Note that if you pick your base of the logarithm to be $10$, $\log10 = 1$ – apnorton Feb 26 '13 at 18:12
This "answer" doesn't really answer the OP's question. Assuming you're working base $10$ then you have $x=25\log_{10}(2)$. Okay, so $x=25y$ where $10^y=2$. We're still left wondering what $y$ is such that $10^y = 2$. – Fly by Night Feb 26 '13 at 18:20

Start with \begin{align} 2 = 10^{\log_{10}(2)} \end{align} and you get \begin{align} 2^{52} = (10^{\log_{10}(2)})^{52}= 10^{52\log_{10}(2)} \end{align}

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$2^{25}$ and not $2^{52}$ :) – hjpotter92 Feb 26 '13 at 18:09
Whoops, yeah that was intendet! His question is left as an exercise ;) – k1next Feb 26 '13 at 18:10

If you want an approximate answer, then use that $2^{10} = 1024 \approx 1000 = 10^3$ and conclude that $2^{25} = (2^{10})^{2.5}\approx (10^{3})^{2.5} = 10^{7.5}$, which is quite close to the actual answer, $10^{7.5257\dots}$.

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