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

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Answer 1 · 2016-05-19T01:44:23

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

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.

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