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Logarithmic equations and inequalities

Logarithmic equations and inequalities

The domain first

An equation with logarithms starts from its domain: every expression inside a logarithm must be positive.

  • At the end, every solution is checked against the domain.
  • A solution outside the domain is no solution of the equation, even if it solves an equation reached on the way.
One logarithm equal to a number

log⁡2(x+1)=3\displaystyle {\log_2(x+1)}={3}. The domain: x>−1\displaystyle x>-1. By the definition of the logarithm:

x+1=23x+1=2^3
  • x+1=8\displaystyle {x+1}={8}, so x=7\displaystyle {x}={7}.
  • 7>−1\displaystyle 7>-1: the solution is in the domain.
A sum of logarithms

log⁡x+log⁡(x−3)=1\displaystyle {\log x+\log(x-3)}={1}. The solution below finds the domain, joins the logarithms by the law of the logarithm of a product, and goes over to the power form.

  • The domain: x>0\displaystyle x>0 and x>3\displaystyle x>3, that is x>3\displaystyle x>3.
  • log⁡(x(x−3))=1\displaystyle {\log(x(x-3))}={1}, so x2−3x=101\displaystyle {x^2-3x}={10^1}.
  • x2−3x−10=0\displaystyle {x^2-3x-10}={0}: x=5\displaystyle {x}={5} or x=−2\displaystyle {x}={-2}.
  • −2\displaystyle -2 is not in the domain. The solution: x=5\displaystyle {x}={5}.
A logarithm equal to a logarithm

Two different numbers cannot have the same logarithm to the same base. So when log⁡af(x)=log⁡ag(x)\displaystyle {\log_a f(x)}={\log_a g(x)}, then f(x)=g(x)\displaystyle {f(x)}={g(x)}.

  • log⁡2(x+3)=log⁡2(2x)\displaystyle {\log_2(x+3)}={\log_2(2x)}. The domain: x>−3\displaystyle x>-3 and x>0\displaystyle x>0, that is x>0\displaystyle x>0.
  • x+3=2x\displaystyle {x+3}={2x}, so x=3\displaystyle {x}={3}, in the domain.
  • log⁡3(x2−2x)=1\displaystyle {\log_3(x^2-2x)}={1} is one of these too: 1=log⁡33\displaystyle {1}={\log_3 3}, so x2−2x=3\displaystyle {x^2-2x}={3}, and x=3\displaystyle {x}={3} or x=−1\displaystyle {x}={-1}. For both x2−2x=3>0\displaystyle {x^2-2x}={3>0}, so both are solutions.
The substitution t=log⁡x\displaystyle {t}={\log x}

In (log⁡x)2−3log⁡x+2=0\displaystyle {(\log x)^2-3\log x+2}={0} the unknown appears only inside log⁡x\displaystyle \log x. We substitute t=log⁡x\displaystyle {t}={\log x} and get a quadratic equation:

t2−3t+2=0t^2-3t+2=0
  • t=1\displaystyle {t}={1} or t=2\displaystyle {t}={2}.
  • log⁡x=1\displaystyle {\log x}={1}: x=10\displaystyle {x}={10}. log⁡x=2\displaystyle {\log x}={2}: x=100\displaystyle {x}={100}.
  • The domain x>0\displaystyle x>0: both solutions are in it.
An inequality between logarithms to the same base

Within the domain:

  • a>1\displaystyle a>1: the function log⁡ax\displaystyle \log_a x increases, so log⁡af(x)>log⁡ag(x)\displaystyle \log_a f(x)>\log_a g(x) where f(x)>g(x)\displaystyle f(x)>g(x) – the sign stays.
  • 0<a<1\displaystyle 0<a<1: the function log⁡ax\displaystyle \log_a x decreases, so log⁡af(x)>log⁡ag(x)\displaystyle \log_a f(x)>\log_a g(x) where f(x)<g(x)\displaystyle f(x)<g(x) – the sign turns over.
  • A number on the other side is written as a logarithm to the same base: 3=log⁡28\displaystyle {3}={\log_2 8}, −2=log⁡124\displaystyle {-2}={\log_{\frac{1}{2}}4}.
  • The answer is the part that the solution and the domain share.
Examples of inequalities
  • log⁡2(x−1)<3\displaystyle \log_2(x-1)<3: the domain x>1\displaystyle x>1. 3=log⁡28\displaystyle {3}={\log_2 8} and the base is greater than 1, so x−1<8\displaystyle x-1<8, that is x<9\displaystyle x<9. With the domain: 1<x<9\displaystyle 1<x<9.
  • log⁡12(x−1)>−2\displaystyle \log_{\frac{1}{2}}(x-1)>-2: the domain x>1\displaystyle x>1. −2=log⁡124\displaystyle {-2}={\log_{\frac{1}{2}}4} and the base is less than 1, so the sign turns over: x−1<4\displaystyle x-1<4, that is x<5\displaystyle x<5. With the domain: 1<x<5\displaystyle 1<x<5.
  • log⁡3(x2−2x)<1\displaystyle \log_3(x^2-2x)<1: the domain x2−2x>0\displaystyle x^2-2x>0, that is x<0\displaystyle x<0 or x>2\displaystyle x>2. x2−2x<3\displaystyle x^2-2x<3 gives −1<x<3\displaystyle -1<x<3. The shared part: −1<x<0\displaystyle -1<x<0 or 2<x<3\displaystyle 2<x<3.
Exercise 1

Solve the equations.

a
    b
      c
        d
          Exercise 2

          Solve the equation.

            Exercise 3

            Solve the inequalities.

            a
              b
                c
                  d
                    e