SOLUTION: 4^(5x+1) = 5^(4x-1) Must solve using logs

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Question 1054592: 4^(5x+1) = 5^(4x-1) Must solve using logs
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
4^(5x+1) = 5^(4x-1) Must solve using logs
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(5x+1)*log(4) = (4x-1)*log(5)
Can you do the rest?

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
4%5E%285x%2B1%29%22%22=%22%225%5E%284x-1%29

Take logs of both sides:

log%284%5E%285x%2B1%29%29%22%22=%22%22log%285%5E%284x-1%29%29 

Move the exponents in front of the logs as multipliers:

%285x%2B1%29log%28%284%29%29%22%22=%22%22%284x-1%29log%28%285%29%29

To make things easier, let single letters stand for
those logs:

Let's substitute the letter A for log(4) and the letter
B for log(5), substituting

log(4) = A and log(5) = B, the equation is simpler as

%285x%2B1%29A%22%22=%22%22%284x-1%29B 

or

A%285x%2B1%29%22%22=%22%22B%284x-1%29

or

5Ax%2BA%22%22=%22%224Bx-B  

Solve for x:

5Ax-4Bx%22%22=%22%22-A-B

%285A-4B%29x%22%22=%22%22-A-B

Divide both sides by 5A+4B

%28%285A-4B%29x%29%2F%285A-4B%29%22%22=%22%22%28-A-B%29%2F%285A-4B%29

%28cross%28%285A-4B%29%29x%29%2Fcross%28%285A-4B%29%29%22%22=%22%22%28-A-B%29%2F%285A-4B%29

x%22%22=%22%22%28-A-B%29%2F%285A-4B%29

Now get your calculator.  If you have a graphing calculator
and know how, you can store log(4) as A and log(5) as B using
the STO key, and type in %28-A-B%29%2F%285A-4B%29 and press ENTER.

Or substitute log(4) for A and log(5) for B

x%22%22=%22%22%28-log%284%29-log%285%29%29%2F%285log%284%29-4log%285%29%29

You'll get x = -6.067672624

Edwin