1. ## Solving Logs

Find the solution of the logarithmic equation
log x + log (x-2) = log (2x)

In terms of logarithms, or correct to four decimal places.
x= _____________

Solve for x:
log (x^5) = (log x)^2

Note there are two solutions, A and B, where A < B
A= ________
B =________

Rewrite the expression:
4 log x - 2 log (x^2 +1) + 5 log (x-1)

as a single logarithm log A. then function
A = _________

2. Originally Posted by B-lap
Find the solution of the logarithmic equation
log x + log (x-2) = log (2x)

In terms of logarithms, or correct to four decimal places.
x= _____________

Solve for x:
log (x^5) = (log x)^2

Note there are two solutions, A and B, where A < B
A= ________
B =________

Rewrite the expression:
4 log x - 2 log (x^2 +1) + 5 log (x-1)

as a single logarithm log A. then function
A = _________
You know the rule $\displaystyle \log a+\log b=\log(ab)$?

So $\displaystyle \log x +\log(x-2)=\log(2x)\implies\log(x^2-2x)=\log(2x)$ $\displaystyle \implies x^2-2x=2x\implies x(x-4)=0\implies x=\{0,4\}$. But $\displaystyle x=0$ doesn't work, because in the original equation, that would make the first term $\displaystyle \log0$, which is undefined. So $\displaystyle x=4$ is the only solution.

Using also the rules that $\displaystyle \log a-\log b=\log\frac{a}{b}$ and $\displaystyle \log(a^b)=b\log a$, see if you can do the others.