SOLUTION: how do you solve log(base4)(4x+9)=1

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Question 1009249: how do you solve log(base4)(4x+9)=1
Found 2 solutions by fractalier, josmiceli:
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
We change this into the exponential form...
log(base4)(4x+9)=1 becomes
4^1 = 4x + 9
4 = 4x + 9
-5 = 4x
x = -5/4

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
It's a trick. Just express +1+ as a log.
In this case: +1+=+log%28+4%2C4+%29+
+log%28+4%2C+%28+4x+%2B+9+%29%29+=+log%28+4%2C4+%29+
Then it follows:
+4x+%2B+9+=+4+
+4x+=+-5+
+x+=+-5%2F4+
----------------
Plug this back into original equation
+log%28+4%2C+%28+4x+%2B+9+%29%29+=+1+
+log%28+4%2C+%28+4%2A%28-5%2F4%29+%2B+9+%29%29+=+1+
+log%28+4%2C+4+%29+=+1+
+1+=+1+
OK