LOGARITHMIC EQUATION

Being the diligent scholar that you are, your mind would have already been infused with the logarithm.

Logarithm, as I told you before, is a way of expressing one member in terms of a base that is raised to some power.

Logarithmic equations are equations where the variable we are solving for appears in the logarithm.

To solve the logarithm equation, there are three things that you must note.

1. The logarithm must be a positive value. This means that the power of a logarithm cannot be negative.

2. If the base of two logarithms is equal to each other, then, we will equate the power.

3. Logarithm can also be converted to an index. This means $\log_{x}y=a$ would be written in index form as $x=y^a$.

Example 1
Find the value of $x$ if 
$\log_{10}x^4+\log_{10}3=\log_{10}768$

Solution:
$\log_{10}3(x^4)=\log_{10}768$

Here, the base is the same, so we equate the logarithm
$3(x^4)=768$

Divide both sides by $3$
$\frac{3(x^4)}{3}=\frac{768}{3}$
$x^4=256$
$\sqrt[\require{cancel}\bcancel{4}]{x^\require{cancel}\bcancel4}$= $\sqrt[4]{256}$ 
$x=\pm{4}$

You know that logarithm can't be negative, hence
$x=4$, $x≠-4$
So, the valid answer is $4$,

Example 2
Solve for $x$ in this equation
$\log_{9}x+\log_{9}(x-3)=\log_{9}10$

Solution:
$\log_{9}x(x-3)=\log_{9}10$
$\log_{9}x^2-3x=log_{9}10$

the base is the same, so we equate the logarithm
$x^2-3x=10$
$x^2-3x-10=0$

Solving quadratically
$x^2-5x+2x-10=0$
$x(x-5)+2(x-5)=0$
$(x+2)(x-5)=0$
$x=-2$ or $x=+5$

In this case, $-2$ is extraneous, hence
$x=5$, x≠$-2$
Therefore, our answer will be $5$

If you are not yet familiar with quadratic equations, refer to this post.

Example 3
Evaluate $\log_{6}x=\frac{9}{\log_{6}x}$

Solution:
Cross multiply
($\log_{6}x)^2=9$

Take the square root of both sides $\sqrt{(\log_{6}x)^2}=\sqrt{9}$
$\log_{6}x=\pm{3}$

This translates to
$\log_{6}x=3$ or $\log_{6}x=-3$

Index form
$x=6^3$ or $x=6^{-3}$
$x=216$ or $x=\frac{1}{216}$

Example 4
Solve for $x$ in the equation
$\log_{7}18+\log_{7}(x-4)=\log_{7}x^2$
 
Solution:
First, we would move all the log with ''$x$'' to one side 
$log_{7}18=\log_{7}x^2-log_{7}(x-4)$
$log_{7}18=\log_{7}\frac{x^2}{x-4}$

The base is the same, so we equate the logarithm.
$18=\frac{x^2}{x-4}$

Cross multiply
$18(x-4)=x^2$
$18x-72=x^2
$0=x^2-18x+72$
$x^2-18x+72=0$
$x^2-6x-12x+72=0$
$x(x-6)-12(x-6)$
$x=12$ or $x=6$
Both answers are both positive values, therefore, they are valid

Example 5
Evaluate 
$\log_{2}(x^2+4x+3)-\log_{2}(x^2+x)=4$

Solution: 
$\log_{2}\frac{x^2+4x+3}{x^2+x}=4$
$\log_{2}\frac{(x+1)(x+3)}{x(x+1)}=4$
$\log_{2}\frac{\require{cancel}\bcancel{(x+1)}(x+3)}{x(\require{cancel}\bcancel{(x+1)}}=4$

Index form
$\frac{x+3}{x}=2^4$
$\frac{x+3}{x}=16$

Cross multiply
$x+3=16(x)$
$x+3=16x$
$3=16x-x$
$3=15x$

Divide both sides by 3
$\frac{3}{15}=\frac{\require{cancel}\bcancel{15}x}{15}$
$\frac{1}{5}=x$
$x=\frac{1}{5}$

Related post
Example 6
Find the value of x if $\log_{4}16^{2x}=\log_{6}216^{x-1}$

Solution:
$\log_{4}4^{2(2x)}=\log_{6}6^{3(x-1)}$
$\log_{4}4^{4x}=\log_{6}6^{3x-3}$
$4x\log_{4}4=(3x-3)\log_{6}6$
$4x(1)=3x-3(1)$
$4x=3x-3$
$4x-3x=-3$
$x=-3$


Example 7
Solve for x in the equation 
$\log_{2}(8x-1)-2\log_{2}(x+1)+\log_{2}(x+4)=3$

Solution:
Recalled that $a\log_{b}c=\log_{b}c^a$
$\log_{2}(8x-1)-\log_{2}(x+1)^2+\log_{2}(x+4)=3$

This translates to
$\log_{2}\frac{(8x-1)(x+4)}{(x+1)^2}=3$

Index form
$\frac{(8x-1)(x+4)}{(x+1)^2}=2^3$
$\frac{(8x-1)(x+4)}{(x+1)^2}=8$
$\frac{8x(x+4)-1(x+4)}{(x+1)(x+1)}=8$

This will translate to
$\frac{8x^2+31-4}{x^2+2x+1}=8$

Let's cross multiply
$8x^2+31x-4=8(x^2+2x+1)$
$8x^2+31x-4=8x^2+16x+8$

Let's collect like the term
$8x^2-8x^2+31x-16x=8+4$
$15x=12$

Divide both sides by $15$
$\frac{\require{cancel}\bcancel{15}x}{\require{cancel}\bcancel{15}}=\frac{12}{15}$
$x=\frac{4}{5}$

That will be all for now. In my next post, we would be trying some killer questions on logarithmic equations, so do well to check it out here.

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