Exponential functions and logarithms: Logarithms
Equations and inequalities with logarithms that have the same base
You can solve the equation \(\log_2(x)=\log_2(x^2+\tfrac{1}{4}\) because the two logarithms can only be equal if their arguments are equal. In this example we have \(x=x^2+\tfrac{1}{4}\), that is \(x^2-x+\tfrac{1}{4}=0\). Factorisation leads to \((x-\tfrac{1}{2})^2=0\) and the solution is \(x=\tfrac{1}{2}\).
This approach works for any equation of the form \(\log_g(A)=\log_g(B)\) provided that you check the final results for correctness. If you started, for example, with the equation \(\log_2(x)=\log_2(x^2-2)\), this approach would lead to the quadratic equation \(x^2-x-2=0\), or \((x+1)(x-2)=0\), with solutions \(x=-1\) and \(x=2\). However \(x=-1\) is not a solution of the original equation because substitution leads to the logarithm of a negative number and this is not allowed because the domain of a logarithmic function consists only of positive real numbers.
An equation with logarithmic functions that have the same base When solving logarithmic equations of the form \(\log_g(A)=\log_g(B)\), with \(A\) and \(B\) arbitrary numbers or mathematical expressions, one uses the following rule: \[\log_g(A)=\log_g(B) \implies A=B\] except that for each solution of the equation \(A=B\) must be checked that substitution in the given equation does not lead to a logarithm with a negative argument.
However, the above rule is not always immediately applicable; often you first have to work towards this form (see the examples below).
\(x={}\) \(\frac{80}{9}\)
The basic rule for logarithms states that if \(\log_g(B)=A\) then \(g^A=B\).
So, with \(g=9, A=2, B=9x+1\) we get: \[9^2=9 x+1\] This means that
\[9 x+1=81\] or \[x=\frac{80}{9}\]
You can also obtain the linear equation \(9^2=9 x+1\) by writing the right-hand side of the equation as a logarithm with base 9 using rules for logarithms: \[\begin{aligned}\log_{9}(9 x+1)&=2\log_{9}(9) \\ &=\log_{9}(9^2)\end{aligned}\] After this you conclude that the arguments of the logarithms on the left- and right-hand side must be equal, i.e., \(9 x+1=9^2\).
A second alternative is to use the left-hand and right-hand side of the given equation as an exponent of a power of 9. In this case: \[9^{\log_{9}(9 x+1)}=9^2\] and the left-hand side is by definition of \(\log_{9}\) equal to \(9 x+1\).
Solving an inequality with logarithms Solving an inequality with logarithms, say \(f(x)<g(x)\), usually involves two steps:
- First solve the corresponding equation, in our example \(f(x)=g(x)\).
- Suppose the inequality consists of continuous functions \(f\) and \(g\) and the solutions of the equation are \(x=x_1,x=x_2, \ldots, x_n\) in ascending order. Then it suffices to check for each of the intervals \((\infty,x_1), (x_1,x_2), \ldots, (x_{n-1},x_n), (x_n,\infty)\) in one point whether the inequality on the interval is true or false.
If one of the functions is a constant, and there is only one solution of the corresponding equation, then it is sufficient to know whether a function is decreasing or increasing.
Of course you can also use the graphs of \(f\) and \(g\), when they are available.
Examples illustrate the above solution method for inequalities with logarithms.
Given is the function \[ f(x) = -4 + \log_{2}(4 x) \] Solve the inequality \(f(x) \lt -3\) exactly.
\(\phantom{xxx}\)
Enter your answer using inequality signs \(\leq\), \(\geq\), \(<\) and \(>\). If the inequality is valid on multiple intervals, separate these using the symbol \(\lor\).
You can also use the formula editor for specifying the answer in interval notation.
The function is defined for \(4 x> 0\), so for \(x > 0\). Thus, the domain of \(f\) is \(D_f = \left(0,\infty\right)\) and the vertical asymptote is the line \(x=0\). If \(x\) is moving towards \(0\) (from the right), \(f(x)\) is moving towards \(-\infty\) and so the function is increasing. To solve the inequality we explore where \(f(x)\) equals \(-3\): \[\begin{aligned} -4 + \log_{2}(4 x) &=-3 \\[0.25cm]
\log_{2}(4 x) &= 1 \\[0.25cm]
4 x &= 2^1 \\[0.25cm]
x &= \frac{1}{2}\end{aligned}\] Since \(f\) an increasing function we conclude that \(f(x)<-3\) for \(0< x<\frac{1}{2}\)
(the function is undefined to the left of the vertical asymptote \(x=0\)).
