SOLUTION: Let x be a real number such that 625^x=64 Then 125x=a√b. What are a and b?

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Question 1204250: Let x be a real number such that 625^x=64 Then
125x=a√b. What are a and b?
Found 3 solutions by MathLover1, ikleyn, MathTherapy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!



Let x+be a real number such that 625%5Ex=64+
Then 125%5Ex=a%2Asqrt%28b%29 What are a+and b?

We have the equation 625%5Ex+=+64
To simplify the equation, we can express both sides with the same base. We know that 625 can be expressed as 5%5E4 and 64 can be expressed as 2%5E6.
Rewriting the equation, we have
%285%5E4%29%5Ex+=+2%5E6
Using the property of exponents, we can simplify further:
5%5E%284x%29+=+2%5E6
To find x, we need to equate the exponents:
4x+=+6

Now, solving for+x:
x+=+6%2F4

Now, we can calculate the value of 125%5Ex using the value of x+we found:
125%5E%286%2F4%29=125%5E%283%2F2%29
since 125=5%5E3, we have %285%5E3%29%5E%283%2F2%29=5%5E%283%283%2F2%29%29=5%5E%289%2F2%29
then
125%5Ex+=+sqrt%285%5E9%29
since sqrt%285%5E9%29=sqrt%28%285%5E4%29%5E2%2A5%29=5%5E4%2Asqrt%285%29=625sqrt%285%29, then
125%5Ex+=+625sqrt%285%29
so, comparing to 125%5Ex=a%2Asqrt%28b%29, we have
a=625
b=5


Answer by ikleyn(52852) About Me  (Show Source):
You can put this solution on YOUR website!
.
highlight%28highlight%28First%29%29,  the problem in the post is written incorrectly.
The correct formulation is as follows 

    Let x be a real number such that 625%5Ex = 64.

    Then  125%5Ex = a%2Asqrt%28b%29.  Find "a" and "b".


highlight%28highlight%28Second%29%29,  the  " solution "  by  @MathLover1 is  TOTALLY  WRONG.
I came to bring a correct solution. 

              highlight%28highlight%28SOLUTION%29%29


We are given  625%5Ex = 64.

It is the same as to write

    5%5E%284x%29 = 64.    (1)


Then

    125%5Ex = 5%5E%283x%29 = %285%5E%284x%29%29%5E%283%2F4%29 = now replace  5%5E%284x%29  by  64,  based on  (1)  = 64^(3/4) = 8^(3/2) = 8%2Asqrt%288%29.


Thus the problem is just solved.  a = 8;  b = 8.     ANSWER



              highlight%28highlight%28CHECK%29%29


If 625^x = 64,  then  x*log(625) = log(64), x = log%28%2864%29%29%2Flog%28%28625%29%29 = 0.646014837  (approximately).


Next,  125%5Ex = 125%5E0.646014837 = 22.627417 (approximately);   

       8%2Asqrt%288%29 = 22.627417  (the same value).   Check is completed and confirms the answer.

Solved.


///////////////////////


For safety of your mind, simply ignore the post by @MathLover1.

Observing her activity at this forum during years, I learned that she DOES NOT know Math, at all
(except of very local pieces).

Her method to work at this forum is to re-write from other tutors or from web-sites
(if she find an appropriate source) - without any reference, naturally.

If she does not find the source to re-write from, she writes any gibberish,
without hesitation.


What others think about her creations, she doesn't care.


Such "tutors" should not be allowed approaching to Math education closer than a cannon shot.



Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be a real number such that 625^x=64 Then
125x=a√b. What are a and b?

That woman is TOTALLY LOST and CLUELESS!!! Who ever heard of EQUATING the EXPONENTS of 2 exponential expressions when they have DIFFERENT bases? 

It appears, as TUTOR @IKLEYN states, that the correct equation is: matrix%281%2C3%2C+125%5Ex%2C+%22=%22%2C+a%2Asqrt%28b%29%29.
                 A different "spin" on this is:  
As seen directly above, 2 is being used as the BASE of 64 i/o 8, as Tutor @IKLEYN did. But, 4 could've also been used
since 64 = 43. So, as you may know and can clearly see, 64 can either be expressed as 82, 26, or 43.
                                                  
                                                       matrix%281%2C3%2C+125%5Ex%2C+%22=%22%2C+matrix%282%2C1%2C+%22+%22%2C+2%5E%289%2F2%29%29%29 ----- EQUATING BASES, since EXPONENTS are the same
                                                       
                                                       matrix%281%2C3%2C+125%5Ex%2C+%22=%22%2C+16sqrt%282%29%29 -------- 
As "a" and "b" are being sought from the equation: 125x = a√b, and matrix%281%2C3%2C+125%5Ex%2C+%22=%22%2C+16sqrt%282%29%29, then IN THIS CASE, a = 16, while b = 2

BTW, 16sqrt%282%29 is the SAME as 8sqrt%288%29. Check this yourself!! And, if base 4 (43) is used for 64, another set of values for "a" and "b" 
- maybe a different set - will ensue! You may want to try that one on your own since 2 of us already used bases 8 and 2. Okay?

Therefore, there is no UNIQUE INTEGER set of values for "a" and "b."