Logarithmic Equations - Examples and Practice Problems

We see that we have a sum of logarithms with the same base on the right-hand side, so we can use the product law to combine them. This is the product law in case you don’t remember it:

Therefore, we have:

$latex \log_{5}(x+1)+\log_{5}(3)=\log_{5}(15)$

$latex \log_{5}[(x+1)3]=\log_{5}(15)$

We can expand the multiplication to get:

$latex \log_{5}(3x+3)=\log_{5}(15)$

The logarithms have the same base, so we can eliminate them and form an equation with the arguments:

$latex 3x+3=15$

The linear equation can be easily solved:

$latex 3x+3=15$

$latex 3x=12$

$latex x=4$

In this case, we have a sum of logarithms on each side of the equation. Therefore, we are going to use the law of the product on both sides to get:

$$\log_{4}(2x+2)+\log_{4}(2)=\log_{4}(x+1)+\log_{4}(3)$$

$latex \log_{4}[(2x+2)2]=\log_{4}[(x+1)3]$

We can expand the multiplication on both sides to get:

$latex \log_{4}(4x+4)=\log_{4}(3x+3)$

Now, we eliminate the logarithms and form an equation with the arguments:

$latex 4x+4=3x+3$

The linear equation can be easily solved:

$latex 4x+4=3x+3$

$latex x=-1$

We can use the product law on the left side:

$$\log_{7}(x)+\log_{7}(x+5)=\log_{7}(2x+10)$$

$latex \log_{7}[x(x+5)]=\log_{7}(2x+10)$

Now, we can distribute the x to get:

$latex \log_{7}({{x}^2}+5x)=\log_{7}(2x+10)$

We can eliminate the logarithm on both sides since they have the same base and we can form an equation with the arguments:

$latex {{x}^2}+5x=2x+10$

We obtained a quadratic equation. We can solve this equation by placing all the terms on one side and factoring the equation:

$latex {{x}^2}+5x=2x+10$

$latex {{x}^2}+5x-2x-10=0$

$latex {{x}^2}+3x-10=0$

$latex (x+5)(x-2)=0$

We solve for each factor:

⇒   $latex x=-5$

⇒   $latex x=2$

Therefore, we have two answers, $latex x=-5$ and $latex x=2$.

In this case, we have log subtractions on both sides of the equation, so we can apply the law of the logarithm quotient. In case you don’t remember, the following is the quotient law:

Therefore, applying this law to both sides, we have:

$$\log_{3}(x+3)-\log_{3}(2)=\log_{3}(x-1)-\log_{3}(7)$$

$latex \log_{3}(\frac{x+3}{2})=\log_{3}(\frac{x-1}{7})$

Expressions within logarithms can no longer be simplified. However, we can eliminate the logarithms since they both have the same base:

$latex \frac{x+3}{2}=\frac{x-1}{7}$

We can cross multiply to simplify:

$latex 7(x+3)=2(x-1)$

We multiply using the distributive property:

$latex 7x+21=2x-2$

We solve the linear equation:

$latex 7x+21=2x-2$

$latex 5x=-23$

$latex x=-\frac{23}{5}$

We can see that the logarithms in this equation do not have a base. When we have logarithms without a base, we assume that the base is 10. Logarithms with base 10 are called common logarithms.

In this equation, we can start by using the power law to rewrite the logarithm that has a fraction in front of it.

Therefore, we have:

$latex \log({{x}^2})+\frac{1}{2}\log(4)=\log({{x}^2}+16)$

$latex \log({{x}^2})+\log({{4}^{\frac{1}{2}}})=\log({{x}^2}+16)$

$latex \log({{x}^2})+\log(2)=\log({{x}^2}+16)$

We simplify the left side using the product law:

$latex \log(2{{x}^2})=\log({{x}^2}+16)$

We can eliminate the logarithm on each side since it has the same base:

$latex 2{{x}^2}={{x}^2}+16$

We can solve for x to solve the quadratic equation:

$latex 2{{x}^2}={{x}^2}+16$

$latex 2{{x}^2}-{{x}^2}=16$

$latex {{x}^2}=16$

Now, we take the square root of both sides:

$latex x=\sqrt{16}$

$latex x=\pm 4$

Therefore, we have two answers, $latex x=4$ and $latex x=-4$.

Here we also don’t have a base in the logarithm, so we know that it is a common logarithm and that its base is 10.

In this equation, we have a one-sided logarithm. This is an equation of the second case mentioned above:

We can solve this equation by writing it in exponential form. Therefore, we remove the logarithm from the left side and write its argument. On the right-hand side, 2 is the exponent and 10 is the base (the base of the logarithm):

$latex \log(4x+60)=2$

$latex 4x+60={{10}^2}$

We apply the exponent and solve the linear equation:

$latex 4x+60=100$

$latex 4x=40$

$latex x=10$

We have to move the logarithms to one side of the equation and the constant terms to the other side:

$latex \log_{2}(3x)-2=\log_{2}(2x-5)$

$latex \log_{2}(3x)-\log_{2}(2x-5)=2$

Now, we simplify the left part using the quotient law:

$latex  \log_{2}(\frac{3x}{2x-5})=2$

To solve, we have to write the equation in its exponential form. The argument remains in the same place and we eliminate the logarithm. We raise the 2 (the base of the logarithm) to the exponent 2:

$latex \frac{3x}{2x-5}={{2}^2}$

$latex \frac{3x}{2x-5}=4$

We simplify by multiplying crosswise:

$latex 3x=4(2x-5)$

We distribute the multiplication:

$latex 3x=8x-20$

We solve the linear equation:

$latex 3x=8x-20$

$latex -5x=-20$

$latex x=\frac{-20}{-5}$

$latex x=4$