SOLUTION: I've tried using logs, but am not sure about actually dividing one log by another (.04)^X *10^9 = 1 .....I think the answer is approximately 30, but can't prove to myself

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I've tried using logs, but am not sure about actually dividing one log by another (.04)^X *10^9 = 1 .....I think the answer is approximately 30, but can't prove to myself      Log On


   

Question 30094: I've tried using logs, but am not sure about actually dividing one log by another

(.04)^X *10^9 = 1 .....I think the answer is approximately 30, but can't prove to myself
Found 2 solutions by josmiceli, Fermat:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
%28.04%29%5Ex+%2A10%5E9+=+1
%284%2F100%29%5Ex+%2A+10%5E9+=+1
4%5Ex+%2A+%281%2F100%29%5Ex+%2A+10%5E9+=+1
take the log(base 10) of both sides
log%284%5Ex%29+%2B+log%28%281%2F100%29%5Ex%29+%2B+9+=+0
(the log of 1 is 0)
x%2Alog%284%29+%2B+x%2A%28-2%29+%2B+9+=+0
(log(a^b) = b*log(a))
x%2A%28log%284%29+-+2%29+=+-9
x+=+-9+%2F+%28.602+-+2%29
x+=+-9+%2F+-1.398
x+=+6.438
to be honest, I don't get the right answer when I plug this
back into %28.04%29%5Ex+%2A10%5E9+=+1
I slipped up somewhere, I think. . . don't know where

Answer by Fermat(136) About Me  (Show Source):
You can put this solution on YOUR website!
0.04%5EX+%2A10%5E9+=+1
take logs (to the base 10) of both sides
log%28%280.04%5EX+%2A+10%5E9%29%29+=+log%281%29
Rule of logs: log%28%28a%2Ab%29%29+=+log%28%28a%29%29+%2B+log%28%28b%29%29 (to any base)
So we get,
log%28%280.04%5EX%29%29+%2B+log%28%2810%5E9%29%29+=+0 (log(1) = 0)
Rule of logs: log%28%28a%5Eb%29%29+=+b%2Alog%28%28a%29%29 (to any base)
So we get,
X+%2A+log%28%280.04%29%29+%2B+9+%2A+log%28%2810%29%29+=+0
X%2Alog%28%280.04%29%29+%2B+9+=+0 (log(10) = 1)
X%2Alog%28%280.04%29%29+=+-9
X+=+%28-9%29%2Flog%28%280.04%29%29
X+=+%28-9%29%2F%28-1.39794%29
X+=+6.4380
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