Let us see the first two methods in action: reduction to the same base and substitution.

Example — Same base

Solve 2x+1=8x12^{x+1} = 8^{x-1}.

Since 8=238 = 2^3, we have 8x1=23(x1)=23x38^{x-1} = 2^{3(x-1)} = 2^{3x-3}. By injectivity of the exponential: x+1=3x3    2x=4    x=2.x + 1 = 3x - 3 \iff -2x = -4 \iff x = 2.

Example — Substitution

Solve 4x32x+2=04^x - 3\cdot 2^x + 2 = 0.

Observe that 4x=(22)x=(2x)24^x = (2^2)^x = (2^x)^2. Setting t=2xt = 2^x with t>0t > 0: t23t+2=0    (t1)(t2)=0    t=1  t=2.t^2 - 3t + 2 = 0 \iff (t-1)(t-2) = 0 \iff t = 1 \ \vee\ t = 2. Both solutions are positive, hence acceptable. Returning to xx:

  • 2x=1=20    x=02^x = 1 = 2^0 \implies x = 0;
  • 2x=2=21    x=12^x = 2 = 2^1 \implies x = 1.

Topics: Exponential function
Concepts: Injectivity
Functions: Exponential function
Methods: Exponential equations
Skills: Solving equations