# Math Help - Solving equation using equality rule for exponents

1. ## Solving equation using equality rule for exponents

${5^(2x-2x^2)}={24*5^-2}+{5^(-2x2)}$

Solve for X

2. Originally Posted by etle1989
${5^(2x-2x^2)}={24*5^-2}+{5^(-2x2)}$

Solve for X
Please fix the formatting so that the equation is clear and unambiguous. Note: Code for things like $5^{2x}$ is 5^{2x}, NOT 5^2x. Failure to use the correct code will result in things like $5^2x$.

3. Originally Posted by etle1989
$5^{2x-2x^2}=24*5^{-2}+5^{-2x^2}$

Solve for X
That's my guess at what you really mean. That " $24*5^{-2}$" really complicates things since we do NOT have just " $5^a= 5^b$

We can rewrite this as $5^{2x- 2x^2}- 5^{-2x^2}= 24*5^{-2}$ which is the same as $5^{2x}5^{-2x2}- 5^{2x^2}= (5^{2x}- 1)(5^{-2x^2})= 24*5^{-2}$. Now, I note that if x= 1, then I have $5^{2x}-1= 25- 1= 24$ and $5^{-2x^2}= 5^{-2}$. However, in general, you cannot just assume that if $ab= 24*5^{-2}$ then $a= 24$ and $b= 5^{-2}$. We can see that x= 1 is a solution but we do not yet know if there are other solutions.

4. Originally Posted by etle1989
$5^{2x-2x^2}=24*5^{-2}+5^{-2x^2}$

Solve for X
1. Re-write the equation:

$\dfrac{5^{2x}}{5^{2x^2}} = \dfrac{24}{25} + \dfrac{1}{5^{2x^2}}$

2. Collecting the variables at the LHS:

$\dfrac{5^{2x}}{5^{2x^2}} - \dfrac{1}{5^{2x^2}}= \dfrac{24}{25}$

$\dfrac{5^{2x}-1}{5^{2x^2}} = \dfrac{24}{25}$

3. That means you have to solve:

$5^{2x}-1 = 24~\wedge~5^{2x^2} = 25$

$5^{2x} = 25 = 5^2~\wedge~5^{2x^2} = 25 = 5^2$

$2x = 2~\wedge~2x^2=2$

which yields x = 1 as the only solution.

5. 1 isn't the only solution.

There are 2 solutions:

$\displaystyle x = 1$

$\displaystyle x \approx 0.238129$

6. Originally Posted by Educated
1 isn't the only solution.

There are 2 solutions:

$\displaystyle x = 1$

$\displaystyle x \approx 0.238129$
Thank you for pointing this out.
Unfortunately I was only able to get a approximate numerical solution:

$5^{2x-2x^2}=24*5^{-2}+5^{-2x^2}$

$5^{2x}=\frac{24}{25}\cdot 5^{2x^2}+1$

$5^{2x}=24\cdot 5^{2x^2-2}+1$

$0=24\cdot 5^{2x^2-2}-5^{2x}+1$

Now using Nerwton's method yields $x \approx 0.2381291343$

7. Originally Posted by earboth
1. Re-write the equation:

$\dfrac{5^{2x}}{5^{2x^2}} = \dfrac{24}{25} + \dfrac{1}{5^{2x^2}}$

2. Collecting the variables at the LHS:

$\dfrac{5^{2x}}{5^{2x^2}} - \dfrac{1}{5^{2x^2}}= \dfrac{24}{25}$

$\dfrac{5^{2x}-1}{5^{2x^2}} = \dfrac{24}{25}$

3. That means you have to solve:

$5^{2x}-1 = 24~\wedge~5^{2x^2} = 25$
this does not follow The fact that $\frac{x}{y}= \frac{a}{b}$ does NOT imply that x= a and y= b!
It does happen to be true for x= 1 but it is not true that we must separate the fractions that way.

$5^{2x} = 25 = 5^2~\wedge~5^{2x^2} = 25 = 5^2$

$2x = 2~\wedge~2x^2=2$

which yields x = 1 as the only solution.

8. $5^{2x}=24\cdot 5^{2x^2-2}+1$

still dont understand how i can simplify that to use equality rule. can someone please explain.

9. Originally Posted by etle1989
$5^{2x}=24\cdot 5^{2x^2-2}+1$

still dont understand how i can simplify that to use equality rule. can someone please explain.
What you want to do cannot be done. The posts in this thread have shown you how the question needs to be solved.