Equations and inequalities with logarithms that have the same base

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.

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).

Find the exact solution of the following equation:

$\log_{4}(4 x-4)=2$

$x={}$ $5$

The basic rule for logarithms states that if $\log_g(B)=A$ then $g^A=B$ .
So, with $g=4, A=2, B=4x-4$ we get:

$4^2=4 x-4$ This means that

$4 x-4=16$ or

$x=5$

You can also obtain the linear equation $4^2=4 x-4$ by writing the right-hand side of the equation as a logarithm with base 4 using rules for logarithms:

$\begin{aligned}\log_{4}(4 x-4)&=2\log_{4}(4) \\ &=\log_{4}(4^2)\end{aligned}$ After this you conclude that the arguments of the logarithms on the left- and right-hand side must be equal, i.e., $4 x-4=4^2$ .

A second alternative is to use the left-hand and right-hand side of the given equation as an exponent of a power of 4. In this case:

$4^{\log_{4}(4 x-4)}=4^2$ and the left-hand side is by definition of $\log_{4}$ equal to $4 x-4$ .

New example

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

$f(x) = -5 + \log_{2}(2 x)$ Solve the inequality $f(x) \lt -4$ 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.

In interval notation $\ivoo{0}{1}$ or as inequality $0<x<1$

The function is defined for $2 x> 0$ , 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 $-4$ :

$\begin{aligned} -5 + \log_{2}(2 x) &=-4 \\[0.25cm] \log_{2}(2 x) &= 1 \\[0.25cm] 2 x &= 2^1 \\[0.25cm] x &= 1\end{aligned}$ Since $f$ an increasing function we conclude that $f(x)<-4$ for $0< x<1$
(the function is undefined to the left of the vertical asymptote $x=0$ ).

New example