SOLUTION: Hi, I tried to solve 5^logx=50-x^log5 in the following manner Re- aligning the equation 5^logx+x^log5 =50 Taking log both sides logx X log5 + logx X log5=log50 log(x+5) + l

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Hi, I tried to solve 5^logx=50-x^log5 in the following manner Re- aligning the equation 5^logx+x^log5 =50 Taking log both sides logx X log5 + logx X log5=log50 log(x+5) + l      Log On


   

Question 478711: Hi,
I tried to solve 5^logx=50-x^log5 in the following manner
Re- aligning the equation
5^logx+x^log5 =50
Taking log both sides
logx X log5 + logx X log5=log50
log(x+5) + log(x+5)=log50
From here am having confusion
1.
log(x+5)(x+5)=log50
log(x+5)^2=log50
(x+5)^2=50=>x=(50)^1/2-(5) (or)
2.
log(x+5) + log(x+5)=log50
2log(x+5)=log50
2(x+5)=50
x+5=25
x=20
Which is correct?
Thanks and regards,
AN
Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
I am sorry, but all incorrect
Your error - you incorrect use formula log%28c%2Ca%29%2Blog%28c%2Cb%29=log%28c%2C+%28ab%29%29
5%5Elogx=50-x%5Elog5
Let t=5%5Elogx
then logt=log%28%285%5Elogx%29%29 Formula log%28c%2Ca%5En%29=n%2Alog%28c%2Ca%29, so
logt=logx%2Alog5 or logt=log5%2Alogx we can write it in this way logt=log%28%28x%5Elog5%29%29, then t=x%5Elog5
So 5%5Elogx=x%5Elog5
Substitute in equation
5%5Elogx=50-5%5Elogx
5%5Elogx%2B5%5Elogx=50
2%2A5%5Elogx=50
5%5Elogx=50%2F2
5%5Elogx=25
5%5Elogx=5%5E2
logx=2
x=10%5E2
x=100