Logarithm help
posted by Joe .
using logarithms to solve exponential equations.
5^1+x = 2^1x
I need exact numbers.
I did one on my own already. 5^x1 = 9
5^x1 = 9
log(5^x1) = log9
(log5)(x1) = log9
x1 = (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^(1x))
(1+x)(log 5)= (1x)(log 2)
(x+1)/(x1)=.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)= (1x)(log 2)
(1+x)/(1x) = log2/log5 = .4306765581
x = 0.39794
That doesn't make sense. How can you jump from 0.43067 to 0.39794?

Logarithm help 
Steve
come on  simple algebra:
(1+x)/(1x) = .4306765581
1+x = .4306765581(1x)
Think you can handle it now?
How about 1+x = 3(1x)
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