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
4 answers
I really need help
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. :)
The answer in the textbook says -0.398. I don't think the answer above gives me -0.398.
(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
1 answer