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How do I write: $$2^x = 5$$

In a logarithmic form?

I've looked for a solution for some time now, so I decided to try here.

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1  
Take log both sides –  Phani Raj May 18 '13 at 10:51
    
If I understand your question correctly, just take the logarithm on both sides to obtain $ \log(2^x) = x \log 2 = \log 5$, and divide by $ \log 2$. –  Henrik Finsberg May 18 '13 at 10:52
4  
Do you know the definition of a logarithm? –  Javier Badia May 18 '13 at 10:53
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5 Answers

up vote 5 down vote accepted

$2^x = 5$

Now 'take log' on both sides:

$\log_2 2^x = \log_2 5$.

We can now use the property $\log a^b = b\times \log a$.

Thus,

$x \times \log_2 2 = \log_2 5$.

But $\log_2 2 = 1$, therefore,

$x = \log_2 5$.

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when you take log on both side is it necessary to use base 2? can we don't use base 5 or any other base ? –  iostream007 May 19 '13 at 12:36
    
@iostream007, you can, but the purpose is to isolate the $x$. For $b^x=y$ you should take $log_b$. –  JMCF125 Jun 15 '13 at 10:10
    
Actually, you can isolate the x with any base. $\log_n a^b = b \log_n a$ –  daviewales Mar 30 at 14:34
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Any equation of the form

$$ \text{base}~^{\textbf{power}} = \text{answer}, $$

can be rewritten as a logarithm as follows:

$$ \log_{~\text{base}}\text{answer} = \textbf{power}. $$

So, $2^x = 5$ quickly becomes $\log_2 5 = x$

Who wants to do all that messing around with logs of both sides and change of base rules?

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There's a simple example that is:

$$\color{green}{10}^\color{red}{3}=\color{blue}{1000}\tag{1}$$

$$\log_{\color{green}{10}} \color{blue}{1000}=\color{red}{3}\tag{2}$$

Notice that the question changed of point:

  • For $(1)$ you have: what's the result of $10$ multiplied by itself $3$ times?

  • For $(2)$ you have: what number should be the exponent of $10$ so that $10^{x}=1000$?

Although Isomorphic's answer is way better because it introduces properties of logarithms, this answer will help you to reflect on the nature of logarithms. I've found this particularly useful on high school.

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Take $\log_e$ then you get $\log_e 2^x=\log_e 5 \implies x=\dfrac{\log_e 5}{\log_e 2}=\log_25$

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you can see all the identity used in other answer in this answer of mine in Write the expressoin in terms of $\log x$ and $\log y \log(\frac{x^3}{10y})$

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