Asked 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
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
Answers
Answered by
Joe
I really need help
Answered by
Savio
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. :)
(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. :)
Answered by
Joe
The answer in the textbook says -0.398. I don't think the answer above gives me -0.398.
Answered by
Steve
(1+x)(log 5)= (1-x)(log 2)
(1+x)/(1-x) = log2/log5 = .4306765581
x = -0.39794
(1+x)/(1-x) = log2/log5 = .4306765581
x = -0.39794
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.