Question 27407: I was wondering if you could perhapse teach me how to solve this question step by step. thank you.
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10^(x-3)=e^(2x+5)
Answer by sdmmadam@yahoo.com(530)   (Show Source):
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10^(x-3)=e^(2x+5)
Given 10^(x-3)=e^(2x+5)
Taking logarithm on both the sides to the base e (the Naperian base)
loge[10^(x-3)] = loge[e^(2x+5)]
which implies (x-3)log10 = (2x+5)loge (the base being e on both the sides)
Here we have taken logarithm on both the sides with an idea to bring the power down using the formula loge(m^n) = n X loge(m)
Therefore (x-3)X loge(10) = (2x+5)X loge(e)
(x-3)y = (2x+5)X 1 (log e to the same base e is 1)
We have put loge(10) = y for convenience.
So, we have (x-3)y = (2x+5)
which is xy -3y = 2x +5
xy - 2x = 5 + 3y (transposing, change side then change sign)
x(y-2) = 5 + 3y
Dividing by (y-2)(which is not zero)
x = (5+3y)/(y-2) where y = loge(10)
Note: (y-2) is not zero because y = loge(10) is not equal to 2
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