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Sladkaya [172]
2 years ago
11

Please help, explain in words the steps needed to solve this problem

Mathematics
1 answer:
melomori [17]2 years ago
4 0

Answer:

x=6

Step-by-step explanation:

By the Product Rule of Logarithms, the logarithm of the product of numbers is the sum of the logarithms of the numbers.

log(x)  +  log(x  - 1)  =  log(x(x  - 1))  =  log( {x}^{2}  -  x )  =  log(3x + 12)

Raise both sides to the power of 10 to eliminate the logs, obtaining:

{x}^{2}  - x = 3x + 12

{x}^{2}  - 4x - 12 = 0

(x - 6)(x  + 2) = 0

x = -2 or x = 6. Since x = -2 is extraneous, it follows that x = 6.

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I need help asap someone please help me i dont understand this question so can someone help me
eduard

Answer:

we conclude that:

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}

Step-by-step explanation:

Given the expression

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}

=\frac{2p}{4p^2-1}\times \frac{6p+3}{6p^3}

=\frac{2p}{4p^2-1}\times \frac{2p+1}{2p^3}

\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}

=\frac{2p\left(2p+1\right)}{\left(4p^2-1\right)\times \:2p^3}

cancel the common factor: 2

=\frac{p\left(2p+1\right)}{\left(4p^2-1\right)p^3}

cancel the common factor: p

=\frac{2p+1}{p^2\left(4p^2-1\right)}

=\frac{2p+1}{p^2\left(2p+1\right)\left(2p-1\right)}

cancel the common factor: 2p+1

=\frac{1}{p^2\left(2p-1\right)}

Expanding

=\frac{1}{2p^3-p^2}

Thus, we conclude that:

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}

7 0
3 years ago
Need help anyone can help
Ket [755]

Answer:

b

Step-by-step explanation:

because

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2 years ago
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If you add 18 to 7% your number comes out as 19.26
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What do you mean by that
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Answer:

Step-by-step explanation:

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