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3 years ago

Rational root theorem

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

8 0

Answer: $a_{n} x^{n} +a_{n-1}x^{n-1}+...+a_{1}x^{1} +a_{0}=0$

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Given the exponential equation 2x = 128, what is the logarithmic form of the equation in base 10?

max2010maxim [7]

Answer:

$log_{2}128= \frac{log_{10} 128}{log_{10} 2}$ .

Step-by-step explanation:

Given : $2^{x}$ = 128.

To find : what is the logarithmic form of the equation in base 10

Solution : We have given that $2^{x}$ = 128.

By th change base rule ( inverse of exponential function )

$a^{b}$ = c is equal to $log_{a}(c)$ = b

Then $2^{x}$ = 128 in to logarithm form $log_{2}(128)$ = x.

Then in to the base 10 logarithmic form.

By the change of base formula $log_{b}a= \frac{log_{10} a}{log_{10} b}$ .

Then , $log_{2}128= \frac{log_{10} 128}{log_{10} 2}$ .

Therefore, $log_{2}128= \frac{log_{10} 128}{log_{10} 2}$ .

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Answer:

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Hope this helped you!

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