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2 years ago

What is the solution to log₂(2x3-8)-2log2x=log2x? O x=1 O x=2 O x=3 O x=4

Mathematics

1 answer:

Artist 52 [7]2 years ago

8 0

The solution to the logarithmic equation as given in the task content is; x = 2.

<h3>What is the solution to the logarithmic expression as given in the task content?</h3>

It follows from the task content that the logarithmic equation as given in the task content is;

log₂(2x3-8)-2log2x=log2x

Hence, upon simplification, we have;

log₂(2x3-8)-log₂x²=log₂x

log₂((2x3-8)/x²)=log₂x

Hence, we have;

(2x³ -8)/x² = x

2x³ -8 = x³

x³ = 8

x = ³√8

x = 2.

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Answer:

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3 years ago

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3 years ago

Read 2 more answers

Inveres laplace transform (3s-14)/s^2-4s+8

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Complete the square in the denominator.

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Rewrite the given transform as

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Now take the inverse transform:

$L^{-1}_t\left\{3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3L^{-1}_t\left\{\dfrac{s-2}{(s-2)^2+2^2}\right\} - 4L^{-1}_t\left\{\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3e^{2t} L^{-1}_t\left\{\dfrac s{s^2+2^2}\right\} - 4e^{2t} L^{-1}_t\left\{\dfrac{2}{s^2+2^2}\right\} \\\\ \boxed{3e^{2t} \cos(2t) - 4e^{2t} \sin(2t)}$

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3 years ago

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