# how to simplify log base 2 and log base 4

How do I simplify the following expression:

$$\log_2(2x+1) - 5\log_4(x^2) + 4\log_2(x)$$

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## 2 Answers

$\log_2(2x+1)-5\log_4x^2+4\log_2x$

$=\log_2(2x+1)+\log_2x^4-5\frac{\log_yx^2}{\log_y4}$

as $\log a+ \log b=\log ab,m\log a=\log a^m$ and $\log_yz=\frac{\log_xz}{\log_xy}$ where $x\neq 1$ as $\log_1y$ is not defined.

$=\log_2(2x+1)x^4-5\frac{\log_yx^2}{\log_y2^2}$

$=\log_2(2x+1)x^4-5\frac{2\log_yx}{2\log_y2}$

$=\log_2(2x+1)x^4-5\log_2x$

$=\log_2(2x+1)x^4-\log_2x^5$

$=\log_2\frac{(2x+1)x^4}{x^5}$

$=\log_2\frac{(2x+1)}{x}$

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Suppose that $x>0$ is some number, and $\log_4x=y$. That means that $4^y=x$. Now $4=2^2$, so $x=4^y=\left(2^2\right)^y=2^{2y}$, and that means that $\log_2x=2y$. In other words, we’ve just demonstrated that for any $x>0$, $\log_2x=2\log_4x$.

Now you have $\log_2(2x+1)-5\log_4x^2+4\log_2x$, which mixes logs base $2$ with logs base $4$; it would be much easier to simplify if all of the logs were to the same base. Use the result of the first paragraph to change $\log_2(2x+1)$ to $2\log_4(2x+1)$ and $\log_2x$ to $2\log_4x$; then you have

$$2\log_4(2x+1)-5\log_4x^2+8\log_4x\;,$$

and you can use the usual properties of logs to express this as the log base $4$ of a single expression.

Going back to $\log_2x=2\log_4x$, if you happen to notice that $2\log_4x=\log_4x^2$, you simply replace $5\log_4x^2$ by $5\log_2x$ to get

$$\log_2(2x+1)-5\log_2x+4\log_2x\;,$$

which is even easier to simplify. The answers that you get by these two approaches won’t be identical, since one will be a log base $4$ and the other a log base $2$, but they’ll be equal, and you can use the relationship $\log_2x=2\log_4x$ to verify this.

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