Solve 2^3x=7^(x+1): Logarithms Solution

[AbsoluteZer0]

Homework Statement

$2^{3x} = 7^{x+1}$

Homework Equations

N/A

The Attempt at a Solution

$2^{3x} = 7^{x+1}$

I'm not entirely sure how to proceed from here. I'm assuming that the following is the right step towards the solution...

$(log2)(3x) = log(7)(x+1)$

...but I'm not entirely sure.

Any suggestions?

Thanks,

 

[Dick]

Homework Statement

$2^{3x} = 7^{x+1}$

Homework Equations

N/A

The Attempt at a Solution

$2^{3x} = 7^{x+1}$

I'm not entirely sure how to proceed from here. I'm assuming that the following is the right step towards the solution...

$(log2)(3x) = log(7)(x+1)$

...but I'm not entirely sure.

Any suggestions?

Thanks,

You are doing just fine. Now solve for x.

 

[AbsoluteZer0]

I've arrived at

$\frac{3x}{x+1} = \frac{log7}{log2}$

I'm not sure how to progress.
What should I do next?

Thanks,

 

[Dick]

I've arrived at

$\frac{3x}{x+1} = \frac{log7}{log2}$

I'm not sure how to progress.
What should I do next?

Thanks,

You are kind of going the wrong way. You've got log(2)*3x=log(7)(x+1)=log(7)x+log(7). Move all the terms involving x to one side and solve for x. log(2) and log(7) are just numbers.