# Guide in Solving Logarithmic Equation

There are some points to remember when solving logarithmic equations. When I was a freshman student, I had a quarrel with my teacher. He gave a quiz about logarithmic equation leading to quadratic.

We know that there are two answers. We need to be very careful with it. Usually only one root will satisfy the original equation and the other one is an extraneous root. I came up with this idea and concluded after a quick check that the other root is extraneous.

So I just shut my mouth up. I know I hate him so much. He is an Engineer, known to be a terror teacher because of number of people failing under his subjects(Algebra and Trigonometry).

I also found out that some of the answers in our examination are incorrect. I tried to approach him again that my answer is correct. He asked me, “How much is your score?” I answered him and again he said, “You must be thankful that you passed, forget it”.

I felt like burning during that time. I worked hard to have a skill in math and he will just mark my answer wrong? Well, he won and I end up with a grade of 8.8/10 in his subject.

Before I reveal that there is always leak in his examination questionnaire, let me give you some points to remember while working with logarithmic equation.

Things to remember when solving logarithmic equation;

1. Domain. The domain can be solved mentally. The purpose of taking note of the domain is to easily determine the extraneous root and the real root without checking.

2. Base of logarithm. If there is more than one logarithmic expression, keep in mind that we can only apply the laws of logarithm if the bases are the same. If in case that the bases are not the same, use the change of base formula to make the bases the same.

3. If the bases are the same, use the laws of logarithm to make a single logarithmic expression in one side of logarithmic equation.

4. Transform is back to its exponential form.

5. Solve for unknown.

6. Make sure to remember these three things. These are simple but it can make magic.

$\log_bb^n=n$ $log_b1=0$ $\log_xy=\displaystyle\frac{\log_by}{\log_bx}$

Worked Problem 1:

Solve for x: $\log_4(3x-4)=4$

Solution:

$\log_4(3x-4)=4$

Rule number 1: Domain is x>4/3

Check rule number 3 and 4.

$\log_2(3x-4)=4 \to 3x-4=2^4$ $3x-4=2^4$ $3x=2^4+4$

$x=\displaystyle\frac{20}{3}$ Since 20/3 is on the domain of logarithm. It is a solution without checking.

Worked Problem 2: 2007 MMC Elimination

Solve for x: $\log_2(x^2-1)-\log_2(x-1)+\log_2(x+5)=5$

Solution:

Domain: x<-1, x>1, x>-5. By quick inspection, the domain must be x>1.

Squeeze the three expressions to a single expression by applying laws of logarithm.

LHS:

$\log_2(x^2-1)-\log_2(x-1)+\log_2(x+5)=5$ $\log_2\displaystyle\frac{(x^2-1)(x+5)}{x-1}=5$

Transform:

$\displaystyle\frac{(x^2-1)(x+5)}{x-1}=2^5$

$\displaystyle\frac{(x-1)(x+1)(x+5)}{x-1}=32$ by factoring.

$(x+1)(x+5)=32$ $x^2+6x+5=32$ $x^2+6x-27=0$ $(x+9)(x-3)=0$ $x=-9,x=3$

The domain is x>1 and only x=3 is on that domain. Which means that x=-9 is an extraneous root.

Worked Problem 3:

Solve for x: $\ln x^2=(\ln x)^2$

Solution:

Take note that $\ln(x)=\log_e(x)$

$\ln x^2=(\ln x)^2$

$2\ln x=(\ln x)(\ln x)$ by dividing lnx both side we can cancel 1 lnx.

$2=\ln x$ $x=e^2$

Worked Problem 4:

Solve for x: $(\ln x)^2+1=2\ln x$

Solution:

Let $a=\ln x$

$a^2+1=2a$ $a^2-2a+1=0$ $(a-1)^2=0$ $a=1$ $\ln x=1$ $x=e^1=e$