Suppose $81^x = 64$. What is $27^{x+1}$?

(3^4)^x = 4^3

3^(4x) = 4^3

3^x = 4^(3/4)

(3^x) ^3 = (4^(3/4))^3
3^(3x) = 4^(3*3/4)
(3^3)^x = 4^(9/4) = (2^2)^(9/4) = 2^(2*9/4) = 2^(9/2)

27^x = 2^(9/2) = 2^(8/2) * 2^(1/2) = 2^4 * 2^(1/2)

27^x = 16 sqrt(2)

27^x * 27 = 16*27 sqrt(2)

27^(x + 1) = 432 sqrt(2)

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xero099 xero099 · Answer 2 · 2021-08-13T21:00:41+02:00

The value of [tex]27^{x+1}[/tex] is approximately 610.146.

In this question, we are going to use Logarithm Properties to determine [tex]x[/tex] in order to calculate the required result. Logarithm Properties are described in the image attached below.

Let [tex]81^{x} = 64[/tex], then the value of [tex]x[/tex] by Logarithm Properties is:

[tex]\log 81^{x} = \log 64[/tex]

[tex]x\cdot \log 81 = \log 64[/tex]

[tex]x = \frac{\log 64}{\log 81}[/tex]

[tex]x \approx 0.946[/tex]

If we know that [tex]x \approx 0.946[/tex], the resulting value is:

[tex]y = 27^{x+1}[/tex]

[tex]y = 27^{1.946}[/tex]

[tex]y \approx 610.146[/tex]

The value of [tex]27^{x+1}[/tex] is approximately 610.146.

Please see the following question for further details: https://brainly.com/question/4314339