Couple of questions dealing with logarithms

I just need to see how they are done, so if you could work them step by step.

If logb3=.234 and logb2=.105, find logb6 .

Solve for x: log(x-5) + log(x-10)=3

438=200e.25x

A=P(1+(r/n))nt for t

4e3x=20

2. This isn't really a calculus problem, but seeing that it's your first post, I'll help you anyway.

For the first problem, we may use the fact that $\displaystyle \log xy = \log x +\log y$. Here,

$\displaystyle \log_b 6 =\log_b 2\cdot 3=\log_b 2+\log_b 3$.

For the second problem, we may deduce a further equality by raising 10 to the power of both sides:

\displaystyle \begin{aligned} 10^{\log(x-5)+\log(x-10)}&=10^3\\ 10^{\log(x-5)}\cdot 10^{\log(x-10)}&=1000\\ (x-5)(x-10)&=1000. \end{aligned}

Two of the remaining problems can be solved by taking the logarithm of both sides.

3. Thanks for the help on the other ones!!

Originally Posted by Scott H
Two of the remaining problems can be solved by taking the logarithm of both sides.

$\displaystyle 438=200e^{0.25x}$

$\displaystyle A=P(1+(r/n))^{nt}$ for $\displaystyle t$

I know I need to use logarithms for these but every time I work them I get a different answer. I just need to see them worked once, then I usually get 'em.

4. $\displaystyle 438=200e^{0.25x} \implies 0.25x = \ln{\frac{438}{200}} \implies x = 4\ln{\frac{438}{200}}$

$\displaystyle A=P(1+(r/n))^{nt} \implies \frac{A}{P}=(1+(r/n))^{nt} \implies \log_{1+(r/n)}\frac{A}{P}=nt \implies$ $\displaystyle t = \frac{\ln{\frac{A}{P}}}{n \cdot \ln({1+(r/n)}) }$