# Logarithms solve for x help

1. Dec 12, 2008

### Draggu

1. The problem statement, all variables and given/known data
Solve for x. State restrictions, if necessary.

3^(2x) - 3^(x) -12 = 0

2. Relevant equations

3. The attempt at a solution

2xlog3 - log3 = log 12
x=log 12/log 3

Doesn't work. I have no idea how to do this.. we didn't learn it. The 2x is throwing me off.

2. Dec 12, 2008

### Avodyne

Re: Logarithms

Let y=3^x, and try to write the equation in terms of y. Then, solve for y. Then, use y=3^x to solve for x.

3. Dec 12, 2008

### Draggu

Re: Logarithms

Let y = 3^(x)

2y-y-12=0
y=12
3^(x)=12
xlog3=12
x = 12/log3

Doesn't work either.

4. Dec 12, 2008

### Avodyne

Re: Logarithms

If y=e^x, what is e^(2x) in terms of y? Hint: the answer is not 2y.

5. Dec 12, 2008

### hoaver

Re: Logarithms

Let $$y=3^x$$
$$y^2-y-12=0$$

$$(y-4)(y+3)=0$$

$$y=4\ or\ y=-3$$

$$3^x=4\ or\ 3^x=-3$$

$$\log 3^x = \log 4$$

$$x= \frac{\log 4}{\log3}$$

$$x \approx 1.2618595071429$$

For any value of x, $$3^x\neq-3$$, this solution is extraneous (rejected).

6. Dec 12, 2008

### Draggu

Re: Logarithms

Are there restrictions btw

7. Dec 12, 2008

### hoaver

Re: Logarithms

Not sure what you mean. The only restriction I'm guessing would be that any negative roots you get will never be a solution to an exponential expression, where the exponent is not 1.