MHA3: Solving Logarithmic Equations

Okay, now we'll solve some logarithmic equations. Log equations are pretty much like your good ol' linear equations, only this time, we'll deal with a bunch of logarithms instead of plain variables. As I said in my previous MHA article, the properties of logarithms are your weapons here, and you'll see why as we solve some examples. Let us list down once again the properties of logarithms.

#1 log_baˣ = xlog_ba
#2 log_b(ax) = log_ba + log_bx
#3 log_b(a/x) = log_ba - log_bx
#4 log_b1 = 0
#5 log_bx = logx / logb

Also, don't forget the statement:
If log_bx = y, then bʸ = x.

Okay. Let's get started!

Example 1:
Solve for x: log_x216 = 3

In this case, we'll just need to recall the good ol' statement about logarithms. We know that the base is x, that the number we're taking the logarithm of to base x is 216, and we know that y, the required exponent, is 3. With these said, we can set up this equation.

x³ = 216 (Take the cubed root of both sides)
x = 6 (Answer)

Example 2:
Solve for x: log₈x = 3

This one's pretty easy! We know that the base is 8, that the number that we're taking the logarithm of to base 8 is x, and that the required exponent is 3. So we can say that:

8³ = x
512 = x (Answer)

Example 3:
Solve for x: log₂x + log₄2x = -1

Solution:
log₂x + log₄2x = -1
(logx/log2) + (log(2x)/log4) = -1 (Property #5)
(logx/log2) + (log(2x)/log(2²)) = -1
(logx/log2) + (log(2x)/2log2) = -1 (Property #1)
(2logx/2log2) + (log(2x)/2log2) = -1
(2logx + log(2x))/2log2 = -1 
(log(x²) + log(2x))/2log2 = -1 (Property #1)
(log(x² * 2x))/2log2 = -1 (Property #2)
(log(2x³))/2log2 = -1
log(2x³) = -2log2
log(2x³) = log(2⁻²) (Property #1)
log(2x³) = log(1/4) (Remove logs)
2x³ = 1/4
x³ = 1/8
x = 1/2 (Answer)

Example 4:
Solve for x: log₃x + log₄x = 4

Here, you'll see that the solution/s in some logarithmic equations aren't integers. But of course you know this. Let's proceed to the solution!

Solution:
log₃x + log₄x = 4
(logx/log3) + (logx/log4) = 4 (Property #5)
([log4logx]/log3log4) + ([log3logx]/log3log4) = 4
(logx4logx + log3logx)/log3log4 = 4

Remember that we can express αx + βx as (α + β)x.
(log4 + log3)(logx) / log3log4 = 4
(log12)(logx) = 4log3log4 (Property #2)
logx = 4log3log4 / log12 
logx = log(3⁴)log4 / log12
logx = log4log81 / log12

Remember that: if log_bx = y, then bʸ = x.
10^(log4log81 / log12) = x
11.6069 ≈ x

You don't really have to follow every single thing that I do here. I'm just doing this to give you ideas on how to solve logarithmic equations. I encourage you to make up your own technique!

Example 5:
Solve for x: log₂(x - 1) = 5 - log₂(x + 3)

Here, we'll refresh our memories about some stuff from our basic algebra; trinomials.
 log₂(x - 1) = 5 - log₂(x + 3)
log₂(x - 1) + log₂(x + 3) = 5
log₂[(x - 1)(x + 3)] = 5 (Property #2)
log₂(x² + 2x - 3) = 5
2⁵ = x² + 2x - 3
32 = x² + 2x - 3
0 = x² + 2x - 3 - 32
0 = x² + 2x - 35 (Factor the trinomial)
0 = (x + 7)(x - 5)
x = -7, 5

BUT! Remember those two things I've told you. You can't take the logarithm of 0, and you can't take the logarithm of a negative number. In this case, x cannot be -7. So the only solution to this problem is x = 5.

Example 6:
Solve for x: ln(x + 3) - ln(x + 2) = 5

"ln" just means "natural logarithm." Natural logarithm is just a fancy name for a logarithm at base e. You probably have seen e, or Euler's number, in your calculator. e is approximately equal to 2.718. With this said, we are actually just dealing with this logarithmic equation: logₑ(x + 3) - logₑ(x + 2) = 5

But since "ln" is easier to type, and since this is the common notation, we'll just stick with this one.
ln(x + 3) - ln(x + 2) = 5
ln[(x + 3)/(x + 2)] = 5 (Property #3)
e⁵ = (x + 3)/(x + 2)
e⁵(x + 2) = x + 3
(e⁵)x + 2e⁵ = x + 3
(e⁵)x - x = 3 - 2e⁵
(e⁵ - 1)x = 3 - 2e⁵
x = (3 - 2e⁵)/(e⁵ - 1)
x ≈ -1.9932

Example 7:
Solve for x: log_x2 + log_√x2 = 9


log_x2 + log_√x2 = 9
(log2/logx) + (log2/log√x) = 9 (Property #5)
(log2/logx) + (log2/logx^(½)) = 9
(log2/logx) + (log2/(1/2)logx) = 9 (Property #1)
(log2/logx) + (2log2/logx) = 9
(2log2 + log2)/logx = 9
3log2/logx = 9
3log2 = 9logx
log(2³) = log(x⁹) (Property #1)
log8 = log(x⁹) (Remove logs)
8 = x⁹
⁹√8 = x
³√(³√8)) = x
³√2 = x (Answer)

I hope these examples gave you an idea about how to solve logarithmic equations. I put some questions here for you readers to answer (if there are any).

1. Solve for x: ln(x + 2) + ln(x - 2) = 3 The answer is √(e^3 + 4).
2. Solve for x: log₃(2x - 1) - 1 = log₃(2x - 3) The answer is 2.
3. Solve for x: log₃p + log₉p = 9/2 The answer is 27.
4. Solve for x: log₂(11 - 6x) = 2log₂(x - 1) + 3 The answer is 3/2.