Solving Exponential Equations

**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