MA001: College Algebra

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers $x > 0$, $S > 0$, $T > 0$ and any positive real number $b$, where $b≠1$,

$log_bS=log_bT$ if and only if $S=T$.

For example,

If  $log_2(x−1)=log_2(8)$, then $x−1=8$.

So, if $x−1=8$, then we can solve for $x$, and we get $x=9$. To check, we can substitute $x=9$ into the original equation: $log_2(9−1)=log_2(8)=3$. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation $log(3x−2)−log(2)=log(x+4)$. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for $x$:

$log(3x−2)−log(2)=log(x+4)$
$log(\dfrac{3x−2}{2})=log(x+4)$	Apply the quotient rule of logarithms.
$\dfrac{3x−2}{2}=x+4$	Apply the one to one property of a logarithm.
$3x−2=2x+8$	Multiply both sides of the equation by 2.
$x=10$	Subtract $2x$ and add 2.

To check the result, substitute $x=10$ into $log(3x−2)−log(2)=log(x+4)$.

$log(\dfrac{28}{2})=log(14)$      The solution checks.

$log_bS=log_bT$ if and only if $S=T$

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

Solve $ln(x^2)=ln(2x+3)$.

$ln(x^2)=ln(2x+3)$
$x^2=2x+3$	Use the one-to-one property of the logarithm.
$x^2−2x−3=0$	Get zero on one side before factoring.
$(x−3)(x+1)=0$	Factor using FOIL.
$x−3=0$ or $x+1=0$	If a product is zero, one of the factors must be zero.
$x=3$ or $x=−1$	Solve for $x$.

There are two solutions: 3 or −1. The solution −1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.