How to write in $2^x=5$ in logarithmic form?
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.
$\endgroup$ 35 Answers
$\begingroup$$2^x = 5$
Now 'take log' on both sides:
$\log_2 2^x = \log_2 5$.
We can now use the property $\log a^b = b\times \log a$.
Thus,
$x \times \log_2 2 = \log_2 5$.
But $\log_2 2 = 1$, therefore,
$x = \log_2 5$.
$\endgroup$ 3 $\begingroup$Any equation of the form
$$ \text{base}~^{\textbf{power}} = \text{answer}, $$
can be rewritten as a logarithm as follows:
$$ \log_{~\text{base}}\text{answer} = \textbf{power}. $$
So, $2^x = 5$ quickly becomes $\log_2 5 = x$
Who wants to do all that messing around with logs of both sides and change of base rules?
$\endgroup$ $\begingroup$There's a simple example that is:
$$\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.
$\endgroup$ $\begingroup$Take $\log_e$ then you get $\log_e 2^x=\log_e 5 \implies x=\dfrac{\log_e 5}{\log_e 2}=\log_25$
$\endgroup$ $\begingroup$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 $\log y \log(\frac{x^3}{10y})$
$\endgroup$More in general
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