how do you solve exponential equation (3/5)^x =2^1-x please I neeed help. THANK you

Hi Bella,

If you take logs of both sides the pesky exponents go away i.e.

Ln((3/5)

^{x}) = Ln(2^{(1-x)})xLn(3/5) = (1-x)Ln(2) = Ln(2) - Ln(2)x

xLn(3/5)+ xLn(2) = Ln(2)

x = Ln(2)/(Ln(3/5)+Ln(2))=3.85

Hope this helps

Jim