SOLUTION: How do i solve this log5^7+1/2log5^4=log5^x

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: How do i solve this log5^7+1/2log5^4=log5^x      Log On


   

Question 473939: How do i solve this log5^7+1/2log5^4=log5^x
Found 2 solutions by ccs2011, MathTherapy:
Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
Use the following log rules:
log(a) + log(b) = log(ab)
log(a) - log(b) = log(a/b)
log(a^n) = n*log(a)
--> log+5%5E7+=+7%2Alog+5
--> %281%2F2%29log+5%5E4+=+%284%29%281%2F2%29log+5+=+2%2Alog+5
--> log+5%5Ex+=+x%2Alog+5
Now the equation looks like this:
7%2Alog+5+%2B+2%2Alog+5+=+x%2Alog+5
Add like terms
9%2Alog+5+=+x%2Alog+5
Divide by log(5) on both sides
9+=+x
Solution is x = 9

Answer by MathTherapy(10837) About Me  (Show Source):
You can put this solution on YOUR website!
How do i solve this log5^7+1/2log5^4=log5^x
*******************************************
The other person got x = 9, but this author disagrees, as this most likely is: 
  log+%285%2C+%287%29%29+%2B+%281%2F2%29log+%285%2C+%284%29%29+=+log+%285%2C+%28x%29%29
 --- Applying a%2Alog+%28b%2C+c%29%29 = log+%28b%2C+%28c%5Ea%29%29
         log+%285%2C+%287%29%29+%2B+log+%285%2C+%282%29%29+=+log+%285%2C+%28x%29%29
                     log+%285%2C+%287%2A2%29%29+=+log+%285%2C+%28x%29%29 ---- Applying log+%28b%2C+%28a%29%29+%2B+log+%28b%2C+%28c%29%29 = log+%28b%2C+%28a%2Ac%29%29
                      log+%285%2C+%2814%29%29+=++log+%285%2C+%28x%29%29
                                 14 = x --- LOG bases are same, so LOG arguments are EQUIVALENT