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

1 Answer
A. S. Adikesavan
May 19, 2016

#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.

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