If log xy 5 and log 3 x 3 5 then find the value of 3y

Video transcript

We're asked to solve the log of x plus log of 3 is equal to 2 log of 4 minus log of 2. So let me just rewrite it. So we have the log of x plus the log of 3 is equal to 2 times the log of 4 minus the log of 2, or the logarithm of 2. And this is a reminder. Whenever you see a logarithm written without a base, the implicit base is 10. So we could write 10 here, 10 here, 10 here, and 10 here. But for the rest of this example, I'll just skip writing the 10 just because it'll save a little bit of time. But remember, this literally means log base 10. So this expression, right over here, is the power I have to raise 10 to to get x, the power I have to raise 10 to to get 3. Now with that out of the way, let's see what logarithm properties we can use. So we know, if we-- and these are all the same base-- we know that if we have log base a of b plus log base a of c, then this is the same thing as log base a of bc. And we also know-- so let me write all the logarithm properties that we know over here. We also know that if we have a logarithm-- let me write it this way, actually-- if I have b times the log base a of c, this is equal to log base a of c to the bth power. And we also know, and this is derived really straight from both of these, is that if I have log base a of b minus log base a of c, that this is equal to the log base a of b over c. And this is really straight derived from these two right over here. Now with that out of the way, let's see what we can apply. So right over here, we have all the logs are the same base. And we have logarithm of x plus logarithm of 3. So by this property right over here, the sum of logarithms with the same base, this is going to be equal to log base 3-- sorry, log base 10-- so I'll just write it here. log base 10 of 3 times x, of 3x. Then, based on this property right over here, this thing could be rewritten-- so this is going to be equal to-- this thing can be written as log base 10 of 4 to the second power, which is really just 16. So this is just going to be 16. And then we still have minus logarithm base 10 of 2. And now, using this last property, we know we have one logarithm minus another logarithm. This is going to be equal to log base 10 of 16 over 2, 16 divided by 2, which is the same thing as 8. So the right-hand side simplifies to log base 10 of 8. The left-hand side is log base 10 of 3x. So if 10 to some power is going to be equal to 3x. And 10 to the same power is going to be equal to 8. So 3x must be equal to 8. 3x is equal to 8, and then we can divide both sides by 3. Divide both sides by 3, you get x is equal to 8 over 3. One way, this little step here, I said, look, 10 to the-- this is an exponent. If I raise 10 to this exponent, I get 3x, 10 to this exponent, I get 8. So 8 and 3x must be the same thing. One other way you could have thought about this is, let's take 10 to this power, on both sides. So you could say 10 to this power, and then 10 to this power over here. If I raise 10 to the power that I need to raise 10 to to get to 3x, well, I'm just going to get 3x. If I raise 10 to the power that I need to raise 10 to to get 8, I'm just going to get 8. So once again, you've got the 3x is equal to 8, and then you can simplify. You get x is equal to 8/3.

As you well know that, a logarithm is a mathematical operation that is the inverse of exponentiation. The logarithm of a number is abbreviated as “log.”

Before we can get into solving logarithmic equations, let’s first familiarize ourselves with the following rules of logarithms:

• The product rule:

The product rule states that the sum of two logarithms is equal to the product of the logarithms.  The first law is represented as;

⟹ log b (x) + log b (y) = log b (xy)

• The quotient rule:

The difference of two logarithms x and y is equal to the ratio of the logarithms.

⟹ log b (x) – log b (y) = log (x/y)

• The power rule:

⟹ log b (x) n = n log b (x)

• Change of base rule.

⟹ log b x = (log a x) / (log a b)

• Identity rule

The logarithm of any positive number to the same base of that number is always 1.
b1=b ⟹ log b (b)=1.

Example:

• The logarithm of the number 1 to any non-zero base is always zero.
  b0=1 ⟹ log b 1 = 0.

How to Solve Logarithmic Equations?

An equation containing variables in the exponents is knowns as an exponential equation. In contrast, an equation that involves the logarithm of an expression containing a variable is referred to as a logarithmic equation.

The purpose of solving a logarithmic equation is to find the value of the unknown variable.

In this article, we will learn how to solve the general two types of logarithmic equations, namely:

1. Equations containing logarithms on one side of the equation.
2. Equations with logarithms on opposite sides of the equal to sign.

How to solve equations with logarithms on one side?

Equations with logarithms on one side take log b M = n ⇒ M = b n.

To solve this type of equations, here are the steps:

• Simplify the logarithmic equations by applying the appropriate laws of logarithms.
• Rewrite the logarithmic equation in exponential form.
• Now simplify the exponent and solve for the variable.
• Verify your answer by substituting it back in the logarithmic equation. You should note that the acceptable answer of a logarithmic equation only produces a positive argument.

Example 1

Solve log 2 (5x + 7) = 5

Solution

Rewrite the equation to exponential form

logs 2 (5x + 7) = 5 ⇒ 2 5 = 5x + 7

⇒ 32 = 5x + 7

⇒ 5x = 32 – 7

5x = 25

Divide both sides by 5 to get

x = 5

Example 2

Solve for x in log (5x -11) = 2

Solution

Since the base of this equation is not given, we therefore assume the base of 10.

Now change the write the logarithm in exponential form.

⇒ 102 = 5x – 11

⇒ 100 = 5x -11

111= 5x

111/5 = x

Hence, x = 111/5 is the answer.

Example 3

Solve log 10 (2x + 1) = 3

Solution

Rewrite the equation in exponential form

log10 (2x + 1) = 3n⇒ 2x + 1 = 103

⇒ 2x + 1 = 1000

2x = 999

On dividing both sides by 2, we get;

x = 499.5

Verify your answer by substituting it in the original logarithmic equation;

⇒ log10 (2 x 499.5 + 1) = log10 (1000) = 3 since 103 = 1000

Example 4

Evaluate ln (4x -1) = 3

Solution

Rewrite the equation in exponential form as;

ln (4x -1) = 3 ⇒ 4x – 3 =e3

But as you know, e = 2.718281828

4x – 3 = (2.718281828)3 = 20.085537

x = 5.271384

Example 5

Solve the logarithmic equation log 2 (x +1) – log 2 (x – 4) = 3

Solution

First simplify the logarithms by applying the quotient rule as shown below.

log 2 (x +1) – log 2 (x – 4) = 3 ⇒ log 2 [(x + 1)/ (x – 4)] = 3

Now, rewrite the equation in exponential form

⇒2 3 = [(x + 1)/ (x – 4)]

⇒ 8 = [(x + 1)/ (x – 4)]

Cross multiply the equation

⇒ [(x + 1) = 8(x – 4)]

⇒ x + 1 = 8x -32

7x = 33 …… (Collecting the like terms)

x = 33/7

Example 6

Solve for x if log 4 (x) + log 4 (x -12) = 3

Solution

Simplify the logarithm by using the product rule as follows;

log 4 (x) + log 4 (x -12) = 3 ⇒ log 4 [(x) (x – 12)] = 3

⇒ log 4 (x2 – 12x) = 3

Convert the equation in exponential form.

⇒ 43 = x2 – 12x

⇒ 64 = x2 – 12x

Since this is a quadratic equation, we therefore solve by factoring.

x2 -12x – 64 ⇒ (x + 4) (x – 16) = 0

x = -4 or 16

When x = -4 is substituted in the original equation, we get a negative answer which is imaginary. Therefore, 16 is the only acceptable solution.

How to solve equations with logarithms on both sides of the equation?

The equations with logarithms on both sides of the equal to sign take log M = log N, which is the same as M = N.

The procedure of solving equations with logarithms on both sides of the equal sign.

• If the logarithms have are a common base, simplify the problem and then rewrite it without logarithms.
• Simplify by collecting like terms and solve for the variable in the equation.
• Check your answer by plugging it back in the original equation. Remember that, an acceptable answer will produce a positive argument.

Example 7

Solve log 6 (2x – 4) + log 6 (4) = log 6 (40)

Solution

First, simplify the logarithms.

log 6 (2x – 4) + log 6 (4) = log 6 (40) ⇒ log 6 [4(2x – 4)] = log 6 (40)

Now drop the logarithms

⇒ [4(2x – 4)] = (40)

⇒ 8x – 16 = 40

⇒ 8x = 40 + 16

8x= 56

x = 7

Example 8

Solve the logarithmic equation: log 7 (x – 2) + log 7 (x + 3) = log 7 14

Solution

Simplify the equation by applying the product rule.

Log 7 [(x – 2) (x + 3)] = log 7 14

Drop the logarithms.

⇒ [(x – 2) (x + 3)] = 14

Distribute the FOIL to get;

⇒ x 2 – x – 6 = 14

⇒ x 2 – x – 20 = 0

⇒ (x + 4) (x – 5) = 0

x = -4 or x = 5

when x = -5 and x = 5 are substituted in the original equation, they give a negative and positive argument respectively. Therefor, x = 5 is the only acceptable solution.

Example 9

Solve log 3 x + log 3 (x + 3) = log 3 (2x + 6)

Solution

Given the equation; log 3 (x2 + 3x) = log 3 (2x + 6), drop the logarithms to get;
⇒ x2 + 3x = 2x + 6
⇒ x2 + 3x – 2x – 6 = 0
x2 + x – 6 = 0……………… (Quadratic equation)
Factor the quadratic equation to get;

(x – 2) (x + 3) = 0
x = 2 and x = -3

By verifying both values of x, we get x = 2 to be the correct answer.

Example 10

Solve log 5 (30x – 10) – 2 = log 5 (x + 6)

Solution

log 5 (30x – 10) – 2 = log 5 (x + 6)

This equation can be rewritten as;

⇒ log 5 (30x – 10) –  log 5 (x + 6) = 2

Simplify the logarithms

log 5 [(30x – 10)/ (x + 6)] = 2

Rewrite logarithm in exponential form.

⇒ 52 = [(30x – 10)/ (x + 6)]

⇒ 25 = [(30x – 10)/ (x + 6)]

On cross multiplying, we get;

⇒ 30x – 10 = 25 (x + 6)

⇒ 30x – 10 = 25x + 150

⇒ 30x – 25x = 150 + 10

⇒ 5x = 160

x = 32