Exponential Equation: 4^x^2 = 2^x

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MF524

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Oct 23, 2008
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I need help solving this exponential equation. My instructions are to find the exact, simplified solution. The problem is

4^x^2=2^x - or 4 to the power of x and x to the power of 2 equals 2 to the power of x.

So far what I have is Step 1: log4(4^x^2)=log4(2x),
Step 2: x^2=log4(2x).

I don't know if I am on the right track please help!!
 
Re: Exponential Equation



Hi, again:

It appears to me that you've overlooked the fact that 4 = 2[sup:2x9r8enq]2[/sup:2x9r8enq].

Are you familiar with the following rule of exponents?

(a^n)^m = a^(n*m)

Rewrite the lefthand side of the original equation expressing 4 as a power of two, and then use the above rule to simplify it.

A solution method should then become obvious.

Cheers,

~ Mark :)

 
Re: Exponential Equation

Thank you for your help Mark! So would the correct method of solving it be as following:

(2^2)^2 = 2^x, which turns into 2^4=2^x, meaning that x= 4?

Did the X on the left hand side disappear because I turned 4 into 2^2 and the exponent 2 was the x? I'm sorry if that didn't make since, I am just trying to make sure I completely understand the problem. Thanks again! Miriam
 
Re: Exponential Equation

MF524 said:
I need help solving this exponential equation. My instructions are to find the exact, simplified solution. The problem is

4^x^2=2^x - or 4 to the power of x and x to the power of 2 equals 2 to the power of x.

So far what I have is Step 1: log4(4^x^2)=log4(2x),
Step 2: x^2=log4(2x).

I don't know if I am on the right track please help!!
Click to expand...

The way you can check your answer is to plug the calculated value/s of 'x' in your original equation and see if it is true for the value.

\(\displaystyle log_4(4^{x^2}) \, = \, log_4(2^x)\)

\(\displaystyle log_4(2^{2x^2}) \, = \, log_4(2^x)\)

\(\displaystyle 2^{2x^2} \, = \, 2^x\)

\(\displaystyle 2x^2 \, = \, x\)

\(\displaystyle 2x^2 \, - \, x \, = \, 0\)

\(\displaystyle x\cdot (2x \, - \, 1) \, = \, 0\)

Now continue....
 
Re: Exponential Equation

so would I continue by doing....
x * (2x/2 - 1/2) = 0, which would mean that X = 1/2 or .5?
 
Re: Exponential Equation



X = 1/2 is one of the solutions.

Look at the last equation in Subhotosh's post.

How would you solve that equation?

(Think: Zero Product Property)

 
Re: Exponential Equation

I'm so sorry but I really cannot figure out how to find any other solution besides x=1/2. Can someone just show me how to work this problem out to the end? I am trying to figure out a problem on a test review for a test I take sunday, and if I can see it worked out I usually understand how to do it again on another problem. Thank you so much for all the help.
 
Re: Exponential Equation

If x(x - a) = 0, then ONE of x or (x - a) must equal 0 ..... kapish?
 
Re: Exponential Equation

If

\(\displaystyle (x \, - \, a)\cdot (x \, - \, b) \, = \, 0\)

then

x = a

or

x = b

you have:


\(\displaystyle (x \, - \, 0)\cdot (x \, - \, \frac{1}{2}) \, = \, 0\)
 
Re: Exponential Equation

Thank you for finding that video on the zero property! It actually helped a lot to see that problem being worked out. I really appreciate everyones help.
 
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