Question

How do you solve `10^(x+1) = 4e^(9-x)`?

Answer 1

`x=(9+ln 4-ln 10)/(1+ln 10)`=2.448, nearly.

Explanation:

Use `ln a^n = n ln a, ln (mn)=ln m + ln n and ln e=log_e e=1`

Equating natural logarithms,

`(x+1) ln 10 = ln(4e^(9-x))`

`=ln 4 + (9-x) ln e`

`= ln 4 + (9-x)(1)`. Solving,

`x=(9+ln 4- ln 10)(1+ln 10)=2.448`, nearly.