SOLUTION: can someone help me with this question thanx. solve the following equation for x, giving the answer correct to two decimal places {{{5^(x-2)=2^(x+3)}}} this question is fro

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: can someone help me with this question thanx. solve the following equation for x, giving the answer correct to two decimal places {{{5^(x-2)=2^(x+3)}}} this question is fro      Log On


   

Question 203500: can someone help me with this question
thanx.
solve the following equation for x, giving the answer correct to two decimal places
5%5E%28x-2%29=2%5E%28x%2B3%29
this question is from the book Longman Pre-U Text STPM Mathematics S & T Paper 1
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
To solve for x when it is in exponents is to use logarithms. Since we want our answer in decimal form (and not logarithm form) we need to use a logarithm we can get from out calculator (or other computing device). If calculators have any logarithms at all they will most likely have base 10 logarithms and/or natural (base e) logarithms. Either will work and give the correct answer to the problem.

5%5E%28x-2%29=2%5E%28x%2B3%29
Find the log of both sides:
log%28%285%5E%28x-2%29%29%29=log%28%282%5E%28x%2B3%29%29%29
Now we can get x out of the exponents by using the following property of logarithms (which applies to all bases): log%28%28a%5Eb%29%29+=+b%2Alog%28%28a%29%29. Applying this to our logarithms we get:
%28x-2%29%2Alog%28%285%29%29=%28x%2B3%29%2Alog%28%282%29%29
%28log%28%285%29%29%29x+-+2log%28%285%29%29+=+%28log%28%282%29%29%29x+%2B+3log%28%282%29%29
We are now in a position to solve for x. We can either
  • replace the logarithms now with decimals and finish the solution with these decimals
  • Or, complete the solution with logarithms and replace them with decimals at the end.

I will use the second approach. But if you can't understand this, then repalce the logarithm's now and finish the solution using the decimals.
Subtract %28log%28%282%29%29%29x from both sides:
%28log%28%285%29%29+-+log%28%282%29%29%29x+-+2log%28%285%29%29+=+3log%28%282%29%29
Add 2log%28%285%29%29 to both sides:
%28log%28%285%29%29+-+log%28%282%29%29%29x+=+3log%28%282%29%29+%2B+2log%28%285%29%29
Divide both sides by %28log%28%285%29%29+-+log%28%282%29%29%29:

We can get the logarithms above from our calculator:

Simplifying gives:
x+=+5.7823539868301501
Rounded to 2 decimal places:
x+=+5.78