SOLUTION: Hi. I've been trying to solve this for about three hours. I know the solution is 11.0834645967, but I can't figure out how to get it within the constraints of the problem: Show h

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Question 136498: Hi. I've been trying to solve this for about three hours. I know the solution is 11.0834645967, but I can't figure out how to get it within the constraints of the problem: Show how to solve the following equation in terns of the common logarithm. Also give an approximation of the solution to the nearest hundreth of a unit. 3^(2x-3)=5^(x+2)
My work is
log(3^2x-3)=log(5^x+2)
(2x-3)log3=(x+2)log5
2x-3=(x+2)log5/log3
(2x-3)/(x+2)=log5/log3
(2x-3)/(x+2)=logbase3 5
and then I get stuck. And I might have gone right past the correct route and gotten more tangled. Any help would be most appreciated.

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Hi. I've been trying to solve this for about three hours. I know the solution is 11.0834645967, but I can't figure out how to get it within the constraints of the problem: Show how to solve the following equation in terns of the common logarithm. Also give an approximation of the solution to the nearest hundreth of a unit.

My work is




You did fine to here.  Then you went astray. Now here is what 
you should have done:

Get your calculator at this point and calculate  as

.

and store it in the memory of your calculator, or write it down.
So you won't have to cary such a long awkward decimal through your
steps, just call it .  In other words let 


Now 

          

becomes simply

          

or switch the A to the left of the parentheses:

          

Distribute on the right:

          

Get all the x terms on the left and other terms on the right:

          

Factor out x on the left:

          

Divide both sides by 

          

Cancel the 's

          

          

Now using the recall feature of your calculator to substitute
the value of A.

          

Now calculate that with your calculator, and you'll get

          

Edwin


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