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how do solve exponential equation

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2 Answers

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
 
 
If you mean (3/5)^x = 2^(1-x). Then take the natural log of both sides.
 
Rule: if y = ax, 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