Solving Logarithmic Equations
Basic Technique for Solving Logarithms

If an equation with logarithms can be solved using algebraic techniques, then those techniques will generally involve the product, quotient, and power rules of logarithms—applied in either direction—as well as examining the problem for common bases. If the equation can be manipulated into the form ${log}_{b}\left(x\right)\=y$ (that is, involving just a single logarithm) then $x\={b}^{y}$.



Example: Solve ${\mathrm{log}}_{4}\left(x\right)\cdot {\mathrm{log}}_{2}\left({x}^{2}\right)\=16$
Solution: First, note that by the power rule ${\mathrm{log}}_{2}\left({x}^{2}\right)\=2{\mathrm{log}}_{2}\left(x\right)$, so the original equation reduces to ${\mathrm{log}}_{4}\left(x\right)\cdot {\mathrm{log}}_{2}\left(x\right)\=8$. Next, using the change of base rule, we have ${\mathrm{log}}_{4}\left(x\right)\=\frac{{\mathrm{log}}_{2}\left(x\right)}{{\mathrm{log}}_{2}\left(4\right)}\=\frac{{\mathrm{log}}_{2}\left(x\right)}{2}$. Substituting this into ${\mathrm{log}}_{4}\left(x\right)\cdot {\mathrm{log}}_{2}\left(x\right)\=8$ and crossmultiplying by 2, we get ${\mathrm{log}}_{2}\left(x\right)\cdot {\mathrm{log}}_{2}\left(x\right)\={\left({\mathrm{log}}_{2}\left(x\right)\right)}^{2}\=16$. Taking the square roots of both sides gives ${\mathrm{log}}_{2}\left(x\right)\=\pm 4$. So, there are two solutions: $x\={2}^{4}\=16$ and $x\={2}^{4}\=\frac{1}{16}$.
Caution: When solving equations involving logarithms, it is very important to keep in mind that the domain of a logarithm function is the positive numbers. As we will see in the examples below, algebraic manipulations of expressions involving logarithms can easily lead to "solutions" which are not valid because of this domain restriction. As a simple illustration, observe that the domain of the function $y\={\mathrm{log}}_{3}\left({x}^{2}\right)$ is $x\ne 0$, while the domain of $y\=2{\mathrm{log}}_{3}\left(x\right)$ is $x\>0$. The product rule for logarithms requires that all the logarithms appearing in the rule be properly defined.

Example


Suppose we want to find the solutions to this equation:
${\mathrm{log}}_{16}\left(x\+13\right)\+{\mathrm{log}}_{16}\left(x2\right)\=1$
Use the (reverse of the) product rule for logarithms:
${\mathrm{log}}_{16}\left(\left(x\+13\right)\cdot \left(x2\right)\right)\=1$
Expand the inner quadratic:
$\phantom{\rule[0.0ex]{0.0em}{0.0ex}}{\mathrm{log}}_{16}\left({x}^{2}\+11\mathbf{}x26\right)\=1$
Each side of the above equation is graphed to the right. The solution to the problem is their intersection. Note that since we can only take the logarithm of positive numbers, the domain of the lefthand side function is the set of all numbers satisfying both $x\+13\>0$ and $x2\>0$, or equivalently $x\>13$ and $x\>2$. This means the domain of the function $y\={\mathrm{log}}_{16}\left(x\+13\right)\+{\mathrm{log}}_{16}\left(x2\right)$ is the set of all numbers satisfying $x\>2$.


Inverting the logarithm by raising 16 to both sides gives:
${x}^{2}+11x26equals;{16}^{1}$
Rewrite this as:
${x}^{2}+11x42equals;0$
This factors as:
$\left(x\+14\right)\cdot \left(x3\right)\=0$, so$xequals;14$ or $x\=3$
Graphed on the right is the function ${x}^{2}\+11x42$, with the two roots shown. However, due to the restrictions placed on the original equation, that is, $x\>2$, only solutions found in the solid region are valid for the original problem. So the solution to the original problem is $x\=3$.

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