Solving an exponential equation

**youngb11** · January 6th 2011, 01:40 PM

I'm a little stuck (again

).

How would I go about solving an equation like this:

$4^2^x-4^x-20=0$

And a more complicated one such as:

$2^x+12(2)^-^x=7$

All help greatly appreciated!

**e^(i*pi)** · January 6th 2011, 01:45 PM

Originally Posted by youngb11

I'm a little stuck (again

).

How would I go about solving an equation like this:

$4^2^x-4^x-20=0$

This is a quadratic in 4^x and happily it will factor to $(4^x-5)(4^x+4)=0$ . Note that $4^x > 0$

And a more complicated one such as:

$2^x+12(2)^-^x=7$

All help greatly appreciated!

Multiply through by the LCD which is 2^x since $2^{-x} = \dfrac{1}{2^x}$ . You'll get $2^{2x} + 12 = 7 \cdot 2^x$ which, again, is a quadratic equation

**Random Variable** · January 6th 2011, 01:46 PM

$(4^{x})^{2} - 4^{x} - 20 = 0$

now let $u = 4^{x}$

**youngb11** · January 6th 2011, 04:38 PM

Originally Posted by e^(i*pi)

Multiply through by the LCD which is 2^x since $2^{-x} = \dfrac{1}{2^x}$ . You'll get $2^{2x} + 12 = 7 \cdot 2^x$ which, again, is a quadratic equation

Thanks! However, I'm a little stuck once we get to $7 \cdot 2^x$ . The answer in the back of the book shows $x=2$ and $x=\dfrac{log3}{log2}$ .

How would we multiply $7$ with $2^x$ but still come to such an answer?

Thanks!

**mr fantastic** · January 6th 2011, 04:44 PM

Originally Posted by youngb11

Thanks! However, I'm a little stuck once we get to $7 \cdot 2^x$ . The answer in the back of the book shows $x=2$ and $x=\dfrac{log3}{log2}$ .

How would we multiply $7$ with $2^x$ but still come to such an answer?

Thanks!

You were told what to do:

Originally Posted by e^(i*pi)

[snip]Multiply through by the LCD which is 2^x since $2^{-x} = \dfrac{1}{2^x}$ . You'll get $2^{2x} + 12 = 7 \cdot 2^x$ which, again, is a quadratic equation

This is not the answer, it is how to get the answer. And how to do this was explained to you with the first question. Let $2^x = w$ . Re-arrange to get a quadratic equation in w that is equal to zero. Solve for w. Hence solve for x.

Please make an atttempt. Show your work, say where you get stuck if you need more help.

**youngb11** · January 6th 2011, 04:46 PM

Originally Posted by mr fantastic

This is not the answer, it is how to get the answer. And how to do this was explained to you with the first question. Let $2^x = w$ . Re-arrange to get a quadratic equation in w that is equal to zero. Solve for w. Hence solve for x.

Please make an atttempt. Show your work, say where you get stuck if you need more help.

Ah, okay. It's much clearer when you replace $2^x$ for W. Thanks again!

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