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(Need asap)Solve each equation by using properties of logs then write as an exponent.


14. Log4 (3x-2x)-log4 (4x+1)=2

Respuesta :

maymayisadayday123

Answer:

Let's solve the equation log₄(3x-2x) - log₄(4x+1) = 2 using the properties of logarithms and then write the result as an exponent.

Given:

log₄(3x-2x) - log₄(4x+1) = 2

Step 1: Simplify the logarithmic expression inside the first parentheses.

log₄(3x-2x) = log₄(x)

Step 2: Rewrite the equation using the simplified expression.

log₄(x) - log₄(4x+1) = 2

Step 3: Use the logarithmic property log₄(a) - log₄(b) = log₄(a/b) to combine the terms on the left side of the equation.

log₄(x/(4x+1)) = 2

Step 4: Apply the exponentiation property of logarithms to both sides of the equation. If log₄(a) = b, then 4ᵇ = a.

4² = x/(4x+1)

Step 5: Simplify the exponent on the left side.

16 = x/(4x+1)

Step 6: Solve the equation for x by cross-multiplying and simplifying.

16(4x+1) = x

64x + 16 = x

64x - x = -16

63x = -16

x = -16/63

Therefore, the solution to the equation log₄(3x-2x) - log₄(4x+1) = 2 is x = -16/63.

Step-by-step explanation:

msm555

Answer:

x = - 18/61 or -0.2951

Step-by-step explanation:

To solve the equation [tex]\sf \log_4(3x-2) - \log_4(4x+1) = 2[/tex], we can use the logarithmic rule for the difference of logarithms:

[tex]\sf \log_b(M) - \log_b(N) = \log_b\left(\dfrac{M}{N}\right)[/tex]

Applying this rule, we rewrite the equation as:

[tex]\sf \log_4\left(\dfrac{3x-2}{4x+1}\right) = 2[/tex]

Now, using the definition of logarithms, we know that if [tex]\sf \log_b(M) = N[/tex], then [tex]\sf b^N = M[/tex].

So, for our equation:

[tex]\sf \dfrac{3x-2}{4x+1} = 4^2[/tex]

[tex]\sf \dfrac{3x-2}{4x+1} = 16[/tex]

Next, we solve for [tex]\sf x[/tex]:

[tex]\sf 3x - 2 = 16(4x + 1)[/tex]

[tex]\sf 3x - 2 = 64x + 16[/tex]

[tex]\sf 3x - 64x = 16 + 2[/tex]

[tex]\sf -61x = 18[/tex]

[tex]\sf x = \dfrac{18}{-61}[/tex]

[tex]\sf x \approx -0.295081967213 [/tex]

[tex]\sf x \approx -0.2951 \textsf{(in 4 d.p.)}[/tex]

Therefore, x = - 18/61 or -0.2951

Note:

Here is a typing mistake:

Question should be: [tex]\sf \log_4(3x-2) - \log_4(4x+1) = 2[/tex]

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