What will the two equation's solutions give?

Question

1.) log₄(x-6) + log₄(x-4) = 4

2.) 6e^6x - 3 = 0

I don't understand it at all, so any explanation provided would be greatly appreciated. Thanks ahead of time!

Answer 1

For the first question, there is a logarithm product rule that states the following:

Log_b(XY) = Log_b(X) + Log_b(Y)

The first equation you asked about is in the form of the right hand side of the above equation. You can condense the left-hand side of your presented equation by following the above Logarithm Product Rule, resulting in Log₄[(x-6)(x+4)]. After some clean up, you would end up at Log₄(x²-2x-24). Thus your resulting equation would be Log₄(x²-2x-24) = 4.

There also exists a logarithm to exponential conversion formula:

Log_b(a) = x can be converted into b^x = a

Using the above conversion, you can take your current Log₄(x²-2x-24) = 4 equation and convert it into 4⁴ = x² - 2x - 24. After cleanup, your equation would be 256 = x² - 2x - 24. The rest is solving a quadratic like you normally would!

For your second question, we can use the conversion formula above to help us again. To more exactly fit the form b^x = a, we must isolate the exponential piece (e^6x). So I would first add three, then divide by 6, resulting in e^6x = 1/2. Now following the conversion formula, we can rewrite this as Log_e(1/2) = 6x. Since you're trying to solve for x, you now just have to divide both sides by 6 and you arrive at x = [Log_e(1/2)]/6 as your final solution!

Hope this helps!

Answer 2

1.) log₄ (x – 6) + log₄ (x – 4) = 4

log₄ (x – 6)(x – 4) = 4 Two logarithms can be combined into a single logarithm when the bases are the same and they are added together

4^{log4 (x - 6)(x - 4)} = 4⁴

(x – 6)(x –4) = 256 The base of the exponent cancels the base of the logarithm

x – 10x + 24 = 256 FOIL the factors

x – 10x – 232 = 0

x = [10 ± √10² –(4)(–232) / 2] Solve the quadratic equation using the quadratic formula

= [(10 ± √100 + 928) / 2]

= [(10 ± √1028) / 2]

= [(10 ± 2√257) /2]

= 5 ± √257

x = 5 + √257 Logarithms cannot be negative so 5 – √257 is not a valid answer

x = 21.0312

Check

log₄ (x – 6) = [log₁₀ (x –6) / log₁₀ 4] Using the logarithm change of base formula

log (21.0312 – 6) = [log₁₀ (21.0312 –6) / log₁₀ 4] = [log (15.0312) / .6021] = 1.1770 / .6021 = 1.9548

log₄ (x – 4) = [log₁₀ (x– 4) / log₁₀ 4]

log₄ (21.0312 – 4) = [log₁₀ (21.0312 – 4) / log₁₀ 4] = [log (17.0312) / .6021] = 1.2312 / .6021 = 2.0449

1.9548 + 2.0449 = 3.9997 = 4

x = 21.0312 or 5 + √257 is the answer

2.) 6e^6x – 3 = 0

6e^6x = 3 add 3 to both sides of the equation

e^6x = 3 / 6 divide both sides by 6

e^6x = 1 / 2 reduce the fraction

log_e (e^6x) = log_e (1 / 2) take the log_e of both sides of the equation

6x = log_e (1 / 2) base of the logarithm cancels the base of the exponent

6x = –.6931

x = –.6931 / 6 divide by sides by 6

x = –.1155

Check

6e^6x – 3 = 0

6e^6(-.1155) –3 = 0

6e^-.6931 – 3 = 0

6(.5) – 3 = 0

3 – 3 = 0

x = –.1155 is the answer

Answer 3

Hello Bobby

The solutions are as follows:

a)

log₄[(x - 6)(x - 4)] = 4, (4)⁴ = (x - 6)(x - 4), 256 = x² - 10x + 24, x² - 10x - 232 = 0.

Using the Quadratic Formula: x = [10 +- sqrt(1028)]/2 = [10 +- 2sqrt(257)]/2 = 5 +- sqrt(257).

Since you cannot have a negative log, the valid solution is: 5 + sqrt(257).

Checking this solution, using Change of Base:

log[5 + sqrt(257) - 6]/log(4) + log[5 + sqrt(257) - 6]/log(4) = 1.95 + 2.05 = 4.

b)

This problem is much easier, i.e., 6e^6x - 3 = 0, 6e^6x = 3, e^6x = 3/6 = 1/2, ln(e)^6x = ln(1/2),

Note ln(e) = 1, 6x = ln(1/2), x = ln(1/2)/6.

If you check this answer, you will see that 3 - 3 = 0.

I hope you understand these solutions.

If you have further questions, please feel free to contact me.

Sincerely

Michael Imbert