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 $\displaystyle \left(\sqrt{4}\right)^x = 2^x$?

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

Re: Values that satisfy exponential equation

Quote:

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 :-)

Re: Values that satisfy exponential equation

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

Re: Values that satisfy exponential equation

Quote:

Originally Posted by

**sa-ri-ga-ma** x^2 - 4x - 3 can be written as x^2 - 4x+ 4 -7. Now factorise.

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

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

Re: Values that satisfy exponential equation

I would use the quadratic formula instead...you should find:

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

Re: Values that satisfy exponential equation

Quote:

Originally Posted by

**MarkFL** I would use the quadratic formula instead...you should find:

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

And, are those the values?

Re: Values that satisfy exponential equation

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

Re: Values that satisfy exponential equation

Quote:

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} - (7^{1/2})^{2}, then factories. Or use quadratic formula to find the factors.