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Logarithms solution

  • #1
125
1

Homework Statement



[itex]2^{3x} = 7^{x+1}[/itex]

Homework Equations



N/A

The Attempt at a Solution



[itex]2^{3x} = 7^{x+1}[/itex]

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

[itex](log2)(3x) = log(7)(x+1)[/itex]

...but I'm not entirely sure.

Any suggestions?

Thanks,
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

Homework Statement



[itex]2^{3x} = 7^{x+1}[/itex]

Homework Equations



N/A

The Attempt at a Solution



[itex]2^{3x} = 7^{x+1}[/itex]

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

[itex](log2)(3x) = log(7)(x+1)[/itex]

...but I'm not entirely sure.

Any suggestions?

Thanks,
You are doing just fine. Now solve for x.
 
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  • #3
125
1
I've arrived at

[itex]\frac{3x}{x+1} = \frac{log7}{log2}[/itex]

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

Thanks,
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
I've arrived at

[itex]\frac{3x}{x+1} = \frac{log7}{log2}[/itex]

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.
 
Last edited:

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