Question 27407
Show for x.
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