Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

Select the correct answer.

What is the solution to this equation?

[tex]216 = 6^{2x - 1}[/tex]

A. [tex]x = 2[/tex]
B. [tex]x = 2.5[/tex]
C. [tex]x = 1[/tex]
D. [tex]x = 3[/tex]

Sagot :

To solve the equation [tex]\(216 = 6^{2x - 1}\)[/tex], we can follow these steps:

1. Take the logarithm of both sides:
To isolate the exponent involving [tex]\(x\)[/tex], take the logarithm base [tex]\(6\)[/tex] of both sides of the equation.
[tex]\[ \log_6(216) = \log_6(6^{2x - 1}) \][/tex]

2. Apply the properties of logarithms:
Use the property of logarithms [tex]\(\log_b(a^c) = c \cdot \log_b(a)\)[/tex] on the right-hand side.
[tex]\[ \log_6(216) = (2x - 1) \cdot \log_6(6) \][/tex]

3. Simplify the logarithm:
We know that [tex]\(\log_6(6) = 1\)[/tex], so the equation simplifies to:
[tex]\[ \log_6(216) = 2x - 1 \][/tex]

4. Solve for [tex]\(x\)[/tex]:
Rearrange the equation to solve for [tex]\(x\)[/tex].
[tex]\[ 2x - 1 = \log_6(216) \][/tex]
Add 1 to both sides:
[tex]\[ 2x = \log_6(216) + 1 \][/tex]
Divide by 2:
[tex]\[ x = \frac{\log_6(216) + 1}{2} \][/tex]

5. Evaluate the logarithm:
Now, evaluate the logarithm [tex]\(\log_6(216)\)[/tex]. The value is approximately 3.

6. Substitute the value:
[tex]\[ x = \frac{3 + 1}{2} = \frac{4}{2} = 2 \][/tex]

Thus, the correct solution to the equation is [tex]\(x = 2\)[/tex]. Therefore, the correct answer is:

A. [tex]\(x = 2\)[/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.