Disguised logarithm equations

A disguised logarithm equation is an equation that can be rewritten in the form of a logarithmic equation, even though it may not look like one at first.

It may help to remind yourself of disguised quadratics, as the process is very similar to solving those.

Solving disguised logarithm equations

The key idea is that we:

• Take an exponential equation, in the form:
  • (2a)^x+a^x = b
• Rewrite it in the form of a quadratic:
  • (a^x)^2 + a^x - b = 0
• Solve the quadratic for a^x, usually by factorising or using the quadratic formula.
• Take the logarithm of both sides to find x.

Solve for x: 9^x + 3^x = 12

• Rewrite 9^x into a quadratic term:
  • =(3^2)^x
  • =3^{2x}
  • =(3^x)^2
• Replace 9^x with (3^x)^2:
  • (3^x)^2 + 3^x = 12
• Rearrange into the form of a quadratic:
  • (3^x)^2 + 3^x - 12 = 0
• Factorise:
  • (3^x + 4)(3^x - 3) = 0
• Set each factor equal to 0:
  • 3^x + 4 = 0 or 3^x - 3 = 0
• Solve each equation using logarithms:
  • For 3^x + 4 = 0:
    • 3^x = -4
    • \log_3(-4) is undefined: no solutions for 3^x=-4
  • For 3^x - 3 = 0:
    • 3^x = 3
    • x = \log_3(3)
    • x = 1
• Answer: x = 1

Solve for x: 4^x + 2^x = 20

• Rewrite 4^x into a quadratic term:
  • =(2^2)^x
  • =2^{2x}
  • =(2^x)^2
• Replace 4^x with (2^x)^2:
  • (2^x)^2 + 2^x = 20
• Rearrange into the form of a quadratic:
  • (2^x)^2 + 2^x - 20 = 0
• Factorise:
  • (2^x + 5)(2^x - 4) = 0
• Set each factor equal to 0:
  • 2^x + 5 = 0 or 2^x - 4 = 0
• Solve each equation using logarithms:
  • For 2^x + 5 = 0:
    • 2^x = -5
    • \log_2(-5) is undefined: no solutions for 2^x=-5
  • For 2^x - 4 = 0:
    • 2^x = 4
    • x = \log_2(4)
    • x = 2
• Answer: x = 2

Solve for x: 16^x + 4^x = 20

• Rewrite 16^x into a quadratic term:
  • =(4^2)^x
  • =4^{2x}
  • =(4^x)^2
• Replace 16^x with (4^x)^2:
  • (4^x)^2 + 4^x = 20
• Rearrange into the form of a quadratic:
  • (4^x)^2 + 4^x - 20 = 0
• Factorise:
  • (4^x + 5)(4^x - 4) = 0
• Set each factor equal to 0:
  • 4^x + 5 = 0 or 4^x - 4 = 0
• Solve each equation using logarithms:
  • For 4^x + 5 = 0:
    • 4^x = -5
    • \log_4(-5) is undefined: no solutions for 4^x=-5
  • For 4^x - 4 = 0:
    • 4^x = 4
    • x = \log_4(4)
    • x = 1
• Answer: x = 1