Asked by Help Please!
solve the exponential function for x using base 5 logarithms.
5^x+4=125^1-3x
5^x+4=125^1-3x
Answers
Answered by
Damon
I suspect you might mean
5^(x+4)=125^(1-3x)
5^(x+4)=(5^3)^(1-3x)
5^(x+4)=5^3(1-3x)
(x+4) = 3 - 9 x
10 x = -1
x = -.1
=====================
or (x+4) log5 (5) = (1-3x) log5(5^3)
(x+4) log5(5) = (1-3x)(3)log5(5)
but log5(5) = 1
5^(x+4)=125^(1-3x)
5^(x+4)=(5^3)^(1-3x)
5^(x+4)=5^3(1-3x)
(x+4) = 3 - 9 x
10 x = -1
x = -.1
=====================
or (x+4) log5 (5) = (1-3x) log5(5^3)
(x+4) log5(5) = (1-3x)(3)log5(5)
but log5(5) = 1
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