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2 years ago

What is the solution to the equation 1/square root of 8 = 4^(m + 2)?

2 answers:

8 0

Take the log (base 4) of both sides of the equation.
$\log_{4}(\frac{1}{\sqrt{8}}) = m + 2$
$\frac{-1}{2} \log_{4}(4^{\frac{3}{2}})-2 = m$
$-2 \frac{3}{4}= m$

The appropriate choice is ...
m = -11/4

salantis [7]2 years ago

6 0

I agree with the other person, the answer is -11/4

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2 years ago

Simplify (8^4/10^4)^-1/4using rational exponents

Vlad1618 [11]

Answer:

$(\frac{8^4}{10^4} )^{-\frac{1}{4} }$ = $\frac{5}{4}$

Step-by-step explanation:

Explanation

Given

$(\frac{8^4}{10^4} )^{-\frac{1}{4} }$

By using algebra formulas

$(\frac{a}{b} )^{m} = \frac{a^m}{b^n}$

$(\frac{8^4}{10^4} )^{-\frac{1}{4} } = \frac{(8^{4})^{\frac{-1}{4} } }{(10^{4} )^{\frac{-1}{4} } }$

$(a^{m} )^{n} = a^{mn}$

$\frac{(8^{4})^{\frac{-1}{4} } }{(10^{4} )^{\frac{-1}{4} } } = \frac{8^{4X\frac{-1}{4} } }{10^{4 X\frac{-1}{4} } }$

= $\frac{8^{-1} }{10^{-1} }$

$= \frac{10}{8} \\= \frac{5}{4}$

Final answer:-

$(\frac{8^4}{10^4} )^{-\frac{1}{4} }$ = $\frac{5}{4}$

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