Solving Exponential Equations with Logarithms: Restrictions and Techniques

[Draggu]

Homework Statement

Solve for x. State restrictions, if necessary.

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

Homework Equations

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.

 

[Avodyne]

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.

 

[Draggu]

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.

Let y = 3^(x)

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

Doesn't work either.

 

[Avodyne]

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

 

[hoaver]

Let y=3^x
<br /> y^2-y-12=0<br />

<br /> (y-4)(y+3)=0<br />

<br /> y=4\ or\ y=-3<br />

<br /> 3^x=4\ or\ 3^x=-3<br />

<br /> \log 3^x = \log 4<br />

<br /> x= \frac{\log 4}{\log3}<br />

<br /> x \approx 1.2618595071429<br />

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

 

[Draggu]

Let y=3^x
<br /> y^2-y-12=0<br />

<br /> (y-4)(y+3)=0<br />

<br /> y=4\ or\ y=-3<br />

<br /> 3^x=4\ or\ 3^x=-3<br />

<br /> \log 3^x = \log 4<br />

<br /> x= \frac{\log 4}{\log3}<br />

<br /> x \approx 1.2618595071429<br />

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

Are there restrictions btw

 

[hoaver]

Are there restrictions btw

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.