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Anon25 [30]
3 years ago
7

If (42)^p = 41^4, what is the value of p?

Mathematics
1 answer:
babymother [125]3 years ago
4 0

First you need to make both bases the same:

Lets remove the ^p and ^4

To make the base of 42 equal to 41, you would have 41^x = 41

X - ln(42) / ln(41) = 1.00648904


Now you have 41^1.00648904(p) = 41^4


Now the bases are equal so we need to set the exponents to equal:

1.00648904(p) = 4

Divide both sides by 1.00648904 to solve for P

P = 4 / 1.00648904

P = 3.97421114

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D

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The augmented matrix for the system of three equaitons is

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\5&2&-2&|&11\\5&-4&4&|&-7\end{array}\right)

Multiply the first row by 5, the second row by -3 and add these two rows:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\5&-4&4&|&-7\end{array}\right)

Subtract the third row from the second:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&6&-6&|&18\end{array}\right)

Divide the third row by 6:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&1&-1&|&3\end{array}\right)

Now multiply  the third equation by 26 and add it to the second row:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&0&-45&|&-90\end{array}\right)

You get the system of three equations:

\left\{\begin{array}{r}3x-4y-5z=-27\\-26y-19z=-168\\-45z=-90\end{array}\right.

From the third equation

z=\dfrac{90}{45}=2.

Substitute z=2 into the second equation:

-26y-19\cdot 2=-168\\ \\-26y-38=-168\\ \\-26y=-168+38=-130\\ \\y=\dfrac{130}{26}=5.

Now substitute z=2 and y=5 into the first equation:

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The solution is (1,5,2)

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2 years ago
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