# Thread: Values that satisfy exponential equation

1. ## Values that satisfy exponential equation

Determine all values of $\displaystyle x$ that satisfy the equation.

i. $\displaystyle x^24^{\frac{x}{2}} - 2x2^{x + 1} - 32^x = 0$

$\displaystyle x^22^x - 2x\cdot2^x - 32^x$

$\displaystyle 2^x\left(x^2 - 4x - 3\right)$

$\displaystyle 2^x\left(x + 2\sqrt{7}\right)\left(x - 2\sqrt{7}\right)$

Values are $\displaystyle x = \pm 2\sqrt{7}$

Also, why can't $\displaystyle 4^{\frac{x}{2}}$ be considered as $\displaystyle \left(\sqrt{4}\right)^x = 2^x$?

2. ## Re: Values that satisfy exponential equation

How have yo got 32^x = 3*2^x. In fact 32^x= (2^5)^x = 2^5x
secondly what you are thinking about ( sqrt4)^x = 2^x is absolutely correct.
please check the question once again if it has been correctly written, in present form we get
2^x[ x^2 - 4x - 2^(4x) ]=0

3. ## Re: Values that satisfy exponential equation

Originally Posted by ibdutt
How have yo got 32^x = 3*2^x. In fact 32^x= (2^5)^x = 2^5x
secondly what you are thinking about ( sqrt4)^x = 2^x is absolutely correct.
please check the question once again if it has been correctly written, in present form we get
2^x[ x^2 - 4x - 2^(4x) ]=0
I noticed that too, but the book gives works with $\displaystyle 32^x$ as $\displaystyle 3\cdot2^x$.
$\displaystyle 4^{\frac{x}{2}}$ is evaluated as $\displaystyle \left(2^2\right)^\frac{x}{2} = 2^x$

I'm trying to figure out the problem along these lines, whether the book is correct or not, I don't know :-)

4. ## Re: Values that satisfy exponential equation

x^2 - 4x - 3 can be written as x^2 - 4x+ 4 -7. Now factorise.

5. ## Re: Values that satisfy exponential equation

Originally Posted by sa-ri-ga-ma
x^2 - 4x - 3 can be written as x^2 - 4x+ 4 -7. Now factorise.
$\displaystyle \left(x^2 - 4x - 3\right)$

$\displaystyle x = \pm\sqrt{7} + 2$

6. ## Re: Values that satisfy exponential equation

$\displaystyle x=2\pm\sqrt{7}$

7. ## Re: Values that satisfy exponential equation

Originally Posted by MarkFL

$\displaystyle x=2\pm\sqrt{7}$
And, are those the values?

8. ## Re: Values that satisfy exponential equation

Yes, as $\displaystyle 2^x=0$ has no solution.

9. ## Re: Values that satisfy exponential equation

Originally Posted by Unreal
$\displaystyle \left(x^2 - 4x - 3\right)$

$\displaystyle \left(x + 2\sqrt{7}\right)\left(x - 2\sqrt{7}\right)$
I did understand how did you get the above factors. In my hint I expected you to proceed by rewriting the expression as
(X -2)2 - (71/2)2, then factories. Or use quadratic formula to find the factors.