# Logarithm help

posted by Joe

using logarithms to solve exponential equations.
5^1+x = 2^1-x
I need exact numbers.

I did one on my own already. 5^x-1 = 9
5^x-1 = 9
log(5^x-1) = log9
(log5)(x-1) = log9
x-1 = (log9/log5)
x= (log9/log5)-1
x = 2.3652
Logarithm help - Joe, Friday, September 18, 2015 at 10:55pm
I really need help
Logarithm help - Savio, Friday, September 18, 2015 at 11:13pm
log(5^(1+x))= log(2^(1-x))
(1+x)(log 5)= (1-x)(log 2)
(x+1)/(x-1)=.4306765581
x+1= .4306765581x-.4306765581

At this point it's a simple algebraic solution to solve for x. :)
Logarithm help - Joe, Friday, September 18, 2015 at 11:27pm
The answer in the textbook says -0.398. I don't think the answer above gives me -0.398.
Logarithm help - Steve, Saturday, September 19, 2015 at 12:27am
(1+x)(log 5)= (1-x)(log 2)
(1+x)/(1-x) = log2/log5 = .4306765581
x = -0.39794

That doesn't make sense. How can you jump from 0.43067 to -0.39794?

1. Steve

come on - simple algebra:

(1+x)/(1-x) = .4306765581
1+x = .4306765581(1-x)

Think you can handle it now?

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