# Solving Exponential Equations

• Dec 2nd 2007, 03:10 PM
Macleef
Solving Exponential Equations
Solve:
$\displaystyle \frac {3}{2}^\frac {m}{2} = \frac {4}{9}$

How do you cross multiply with this equation?

------

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?

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

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

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

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

$\displaystyle x + 1 = -2$

$\displaystyle x = -3$

-------------------------------------------

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

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

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

$\displaystyle 2^a = 2^5$

$\displaystyle a = 5$

----------------------------------------

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

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

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

$\displaystyle m = 5$

----------------------------------------------

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

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

$\displaystyle 5^n = 5^5$

$\displaystyle n = 2$

• Dec 2nd 2007, 04:36 PM
topsquark
Quote:

Originally Posted by Macleef
Solve:
$\displaystyle \frac {3}{2}^\frac {m}{2} = \frac {4}{9}$

How do you cross multiply with this equation?

------

Is this possibly supposed to be
$\displaystyle \left ( \frac{3}{2} \right )^{m/2} = \frac{4}{9}$

Then note that
$\displaystyle \frac{4}{9} = \left ( \frac{2}{3} \right )^2 = \left ( \frac{3}{2} \right )^{-2}$

So
$\displaystyle \left ( \frac{3}{2} \right )^{m/2} = \left ( \frac{3}{2} \right )^{-2}$

Thus
$\displaystyle \frac{m}{2} = -2$

$\displaystyle m = -4$

-Dan
• Dec 2nd 2007, 04:39 PM
topsquark
Quote:

Originally Posted by Macleef
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?

-----

You have the right strategy:
$\displaystyle \frac {27^x}{9^{2x-1}} = 3^{x+4}$

$\displaystyle \frac {(3^3)^x}{(3^2)^{2x-1}} = 3^{x+4}$

$\displaystyle \frac {3^{3x}}{3^{2(2x-1)}} = 3^{x+4}$

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

$\displaystyle 3^{3x - (4x - 2)} = 3^{x+4}$

$\displaystyle 3^{-x + 2} = 3^{x+4}$

You take it from here.

-Dan
• Dec 2nd 2007, 04:40 PM
topsquark
Quote:

Originally Posted by Macleef
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: $\displaystyle 10^z = \frac {1}{9900}$

There's a baaaad typo in here somewhere.

-Dan
• Dec 2nd 2007, 04:45 PM
topsquark
Quote:

Originally Posted by Macleef
Solve:
$\displaystyle 27(3^{3x+1}) = 3$

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

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

$\displaystyle x + 1 = -2$

$\displaystyle x = -3$

There's a typo in the third line of your solution.

Quote:

Originally Posted by Macleef
Solve:
$\displaystyle 2^{a+5} + 2^a = 1056$

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

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

$\displaystyle 2^a = 2^5$

$\displaystyle a = 5$

-Dan
• Dec 2nd 2007, 04:49 PM
topsquark
Quote:

Originally Posted by Macleef
Solve:
$\displaystyle 3^{m+1} + 3^{m+2} - 972 = 0$

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

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

$\displaystyle m = 5$

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

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

$\displaystyle 3^{m + 1} ( 1 + 3^1 ) = 972$ <-- See this line

$\displaystyle 3^{m + 1}(4) = 972$

$\displaystyle 3^{m + 1} = 243 = 3^5$

etc.

-Dan
• Dec 2nd 2007, 04:54 PM
topsquark
Quote:

Originally Posted by Macleef

Solve
:

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

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

$\displaystyle 5^n = 5^5$

$\displaystyle n = 2$

Again, I'm not agreeing with the textbook.

A couple of problems here:
$\displaystyle 5^{n+2} - 5^{n+3} = -2500$

$\displaystyle 5^{n+2}(1 - 5^1 ) = -2500$

$\displaystyle 5^{n + 2}(-4) = -2500$

$\displaystyle 5^{n + 2} = 625 = 5^4$

etc.

$\displaystyle 5^n (5^5 - 5^3) = -12500$
It's fine to merely pull the $\displaystyle 5^n$ out, but you should have
$\displaystyle 5^n(5^2 - 5^3) = -2500$