Solve:

$\displaystyle \frac {3}{2}^\frac {m}{2} = \frac {4}{9} $

How do you cross multiply with this equation?

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Solve:

$\displaystyle \frac {27^x}{9^{2x-1}} = 3^{x+4}$

What do you do here? The first thing I did is putting everything to a common base and distributing the exponents then what?

Textbook Answer:-1

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Solve:

$\displaystyle 10^{z+4} = 10^{z+3} = 11$

Do you move $\displaystyle 10^{z+3}$ to the right and solve? I did this and I get a different answer from my textbook.

My answer: $\displaystyle 10^z = \frac {1}{9900}$

Textbook answer: -3

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Solve:

$\displaystyle 27(3^{3x+1}) = 3 $

My Work:

$\displaystyle 3^{3x+1} = \frac {1}{9}$

$\displaystyle 3^{3x+1} = 3^{-2}$

$\displaystyle x + 1 = -2$

$\displaystyle x = -3$

Textbook Answer:-1

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Solve:

$\displaystyle 2^{a+5} + 2^a = 1056$

My Work:

$\displaystyle 2^a (2^5 + 2^0) = 1056$

$\displaystyle 2^a(33) = 1056$

$\displaystyle 2^a = 2^5$

$\displaystyle a = 5$

Textbook Answer: -1

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Solve:

$\displaystyle 3^{m+1} + 3^{m+2} - 972 = 0$

My work:

$\displaystyle 3^{m+1} (3^0 + 3^{-1}) = 972$

$\displaystyle 3^{m+1} = 3^6$

$\displaystyle m = 5$

Textbook answer: -2

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Solve:

$\displaystyle 5^{n+2} - 5^{n+3} = -2500$

My Work:

$\displaystyle 5^n (5^5 - 5^3) = -12500$

$\displaystyle 5^n = 5^5$

$\displaystyle n = 2$

Textbook Answer: 4