# Exponents with different bases

1. Jan 14, 2009

### Mentallic

1. The problem statement, all variables and given/known data
This is re-asking the question in Logarithms! which mysteriously went quiet before the answer could be given... Anyway I also tried answering the question, but to no avail.

2. Relevant equations
$$a^m.a^n=a^{m+n}$$
$$(a^m)^n=a^{mn}$$
$$log(ab)=loga+logb$$
$$log(a^b)=bloga$$

3. The attempt at a solution
$$12^x=4.8^{2x}$$
$$3^x.(2^2)^x=2^2(2^3)^{2x}$$
$$3^x.2^{2x}-2^{6x+2}=0$$
$$2^{2x}(3^x-2^{4x-2})=0$$

Hence, $$2^{2x}=0$$ (1) or $$3^x-2^{4x-2}=0$$ (2)
(1) has no solution.

(2) $$\Rightarrow$$ $$3^x=2^{4x+2}$$
$$xlog3=(4x+2)log2$$

How can one go about solving this? different bases? It gives an answer of $$x\approx -0.82814$$

2. Jan 14, 2009

### Mentallic

Oh hang on...

$$xlog3=(4x+2)log2$$
$$xlog3=4xlog2+2log2$$
$$x(log3-4log2)=2log2$$
$$x=\frac{2log2}{log3-4log2}$$

$$x\approx -0.82814449$$

Ahh never mind.

3. Jan 14, 2009

### rosh300

12^x = 4X8^(2x)
take logs of both sides: log(12x) = log(4 X 82x)
Get rid of the power 2 (althought not essential) log(12x) = log(4 X 64x)
Split RHS using the laws of logs and bring the powers out: x log(12) = log(4) + x log(64)
see if u can do the last step (hint collect all terms with x)

4. Jan 14, 2009

### Дьявол

It would rather use log3.