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Exponential equations and inequalities

Exponential equations and inequalities

The same base on both sides

In an exponential equation the unknown is in the exponent. When both sides are powers of the same base, the exponents are equal:

af(x)=ag(x)  ⟺  f(x)=g(x)\displaystyle a^{f(x)}=a^{g(x)}\iff f(x)=g(x)a>0, a≠1\displaystyle a>0,\ a\neq1
  • 9x=27\displaystyle {9^{x}}={27}: 32x=33\displaystyle {3^{2x}}={3^{3}}, so 2x=3\displaystyle {2x}={3} and x=32\displaystyle {x}={\frac{3}{2}}.
  • 2x=−4\displaystyle {2^{x}}={-4} has no solution: a power of a positive base is always positive.
A common factor

When the exponents differ by a constant number, take the smaller power out as a common factor. The example below solves 2x+1+2x=24\displaystyle {2^{x+1}+2^{x}}={24}:

  • 2x+1=2⋅2x\displaystyle {2^{x+1}}={2\cdot2^{x}}, so the left side is 2⋅2x+2x=3⋅2x\displaystyle {2\cdot2^{x}+2^{x}}={3\cdot2^{x}}.
  • 3⋅2x=24\displaystyle {3\cdot2^{x}}={24}, so 2x=8=23\displaystyle {2^{x}}={8}={2^{3}} and x=3\displaystyle {x}={3}.
The substitution t=ax\displaystyle {t}={a^x}

When an equation holds ax\displaystyle a^x and a2x=(ax)2\displaystyle {a^{2x}}={(a^x)^2}, substitute t=ax\displaystyle {t}={a^x} and you get a quadratic equation. The example below solves 4x−3⋅2x+1+8=0\displaystyle {4^{x}-3\cdot2^{x+1}+8}={0}:

  • 4x=(2x)2=t2\displaystyle {4^{x}}={(2^{x})^{2}}={t^{2}}, and 2x+1=2t\displaystyle {2^{x+1}}={2t}.
  • t2−6t+8=0\displaystyle {t^{2}-6t+8}={0}, so t=2\displaystyle {t}={2} or t=4\displaystyle {t}={4}.
  • 2x=2\displaystyle {2^{x}}={2} gives x=1\displaystyle {x}={1}; 2x=4\displaystyle {2^{x}}={4} gives x=2\displaystyle {x}={2}.
  • t=ax\displaystyle {t}={a^x} is always positive: a negative solution for t\displaystyle t gives no x\displaystyle x.
The same exponent, different bases

Powers with the same exponent join into one power by ax⋅bx=(a⋅b)x\displaystyle {a^x\cdot b^x}={(a\cdot b)^x}:

  • 2x⋅3x=36\displaystyle {2^{x}\cdot3^{x}}={36}: 6x=62\displaystyle {6^{x}}={6^{2}}, so x=2\displaystyle {x}={2}.
  • 3x=5x\displaystyle {3^{x}}={5^{x}}: divide by 5x\displaystyle 5^x and you get (35)x=1\displaystyle {\left(\frac{3}{5}\right)^{x}}={1}, so x=0\displaystyle {x}={0}.
Exponential inequalities

ax\displaystyle a^x increases when a>1\displaystyle a>1 and decreases when 0<a<1\displaystyle 0<a<1. So when you compare the exponents:

  • for a>1\displaystyle a>1 the sign stays: 2x>25\displaystyle 2^{x}>2^{5} means x>5\displaystyle x>5;
  • for 0<a<1\displaystyle 0<a<1 the sign turns: (13)x>(13)2\displaystyle \left(\frac{1}{3}\right)^{x}>\left(\frac{1}{3}\right)^{2} means x<2\displaystyle x<2.
Worked example: an inequality by substitution

The example below solves 9x−10⋅3x+9<0\displaystyle 9^{x}-10\cdot3^{x}+9<0 with the substitution t=3x\displaystyle {t}={3^x}:

  • t2−10t+9<0\displaystyle t^{2}-10t+9<0, that is (t−1)(t−9)<0\displaystyle (t-1)(t-9)<0, so 1<t<9\displaystyle 1<t<9.
  • 1<3x<9\displaystyle 1<3^{x}<9 is 30<3x<32\displaystyle 3^{0}<3^{x}<3^{2}, and the base 3 is greater than 1.
  • The solution: 0<x<2\displaystyle 0<x<2.
Exercise 1

Solve.

a
    b
      c
        d
          Exercise 2

          Solve with a common factor.

          a
            b
              c
                Exercise 3

                Solve with a substitution.

                a
                  b
                    c
                      Exercise 4

                      Solve the inequalities.

                      a
                        b
                          c
                            d