Logarithm (Exponential)

Question

f(x) = 8e^3x + 27e^-3x . By using the substitution z= e^3x solve, for x > 0, the equation f(x) = 35.

Answer 1

f(x) = 8e^3x + 27e^-3x

f(x) = 8e^3x + 27/e^3x (since a^-b = 1/a^b)

 

Now, substitute z=e^3x and f(x) = 35

35 = 8z + 27/z

Multiply each term with z

35z = 8z² + 27

8z² - 35z + 27 = 0

(z-1)(8z-27) = 0

z=1, or z=27/8

Now, we need to find out which of these 2 values of z are valid.

Since z=e^3x and our solution x needs to satisfy the condition x>0

=> z>1

Therefore, z=1 is not valid and the only answer is z=27/8

 

Now, substitute z=e^3x, to solve for x

e^3x = 27/8

ln(e^3x) = ln(27/8)

3x = ln(27/8)

x = (1/3)*ln(27/8) = 0.405