how do solve exponential equation

Question

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

 

Answer 1

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

 

 

Answer 2

If you mean (3/5)^x = 2^(1-x). Then take the natural log of both sides.

 

Rule: if y = a^x, then  ln(y) = x ln(a)

 

x ln(3/5) = (1-x) ln 2

 

x ln (3/5) = ln 2 - x ln 2

 

x (ln (3/5) + ln 2) = ln 2

 

x = ln 2 / [ln (3/5) + ln 2]

 

x = 3.8