Home : Support : Online Help : Math Apps : Functions and Relations : Logarithmic Functions : MathApps/SolvingLogarithmicEquations
var worksheet = new Worksheet();
function doSubmit() {
worksheet.updateComponents();
return true;
}
function reexecuteWorksheet() {
worksheet.reexecuteWorksheet();
}
function refreshWorksheet() {
worksheet.refreshWorksheet();
}
function reloadWorksheet() {
worksheet.reloadWorksheet();
}
function closeWorksheet() {
worksheet.closeWorksheet();
}
function stopWorksheet() {
worksheet.stopWorksheet();
}
function pingWorksheet() {
worksheet.pingWorksheet();
}
var tempComponent;
worksheet.addComponent('table250_ecmath275', new ECMath('table250_ecmath275'));
worksheet.addComponent('table250_ecmath293', new ECMath('table250_ecmath293'));
worksheet.addComponent('table250_ecmath311', new ECMath('table250_ecmath311'));
worksheet.addComponent('table250_ecmath329', new ECMath('table250_ecmath329'));
worksheet.addComponent('table250_ecmath347', new ECMath('table250_ecmath347'));

Solving Logarithmic Equations
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 cross-multiplying 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.
${}$
${}$ ${}$ |

## Was this information helpful?

## Tell us what we can do better:

What kind of issue would you like to report?

(Optional)

(Optional)