It would be great if you could delineate, your play with algebra . So that member could understand it!

Regards
S

we have (4/10)x−1=(625/100)6x−5 try to take RHS and look at those fraction in bracket , find something similar , oh yes , reduce the fraction (625/100) to (25/4) or you can write it like (5/2)^2 correct ? now take the LHS and reduce the fraction of (4/10) to (2/5) easy right ? Oh yes , so what now ? okay so now we have the equation in reduce form which looks like (2/5)^x-1 = (5/2)^2*(6x-5) i hope you still remember why do we have multiplication of 2 in power . now we can write (2/5)^x-1 to (5/2)-(x-1) i hope you remember the property that (a/b) = (b/a)^-1 they both are same thing and thats what i have done . now we have (5/2)^-x+1 = (5/2)^12x-10 now can equate the powers as we have same bases
therefore we have -x+1= 12x-10
and x = 13/11
put the value for x and check yourself
thankyou !

I reached the following step but don't know how to proceed further.

x=11/13

option A: (13/11)^2

option B: (2)^(13/11)

How to proceed further to compare the two values?

One simple way is:

\(\frac{11}{13} = 0.84\)
And, we know \(\sqrt{2} = 1.414\)

So, \(\sqrt{2} < 2^{0.84} < 2^1\)

Col. A: \(\frac{169}{121} = 1.396\)
Col. B: \(2^{0.84}\)

Hence, Col. A < Col. B

Thank you KarunMendiratta. Kudos!

KarunMendiratta, can you please explain what approach should be taken to solve this question under time constraint?

LHS

\((\frac{4}{10})^x*(\frac{4}{10})^{-1}=(\frac{4}{10})^x*(\frac{10}{4})=\frac{2^{2x}}{2}=\)

Basically the LHS is 2x-1

RHS

\((\frac{625}{100})^{6x}*(\frac{100}{625})^5=\)

Basically it tells us that 6x-5

\(2x-1=6x-5
\)

\(-4x=-4\)

\(x=1\)

Substitute

B is the answer

Carcass, thank you for your answer above. Can you please explain on what basis you equated powers in the LHS and RHS? My understanding is that we can only equate powers when bases in two sides are equal. In this case, bases in the right and left hand sides are not equal.

In addition, you simplified LHS to the following:
(4/10)x∗(10/4)=2^2x/2. Why did you eliminate 10?

This took me less than 90 seconds to solve, hope you can do it too.

\((\frac{4}{10})^{x - 1} = (\frac{625}{100})^{6x - 5}\)

\((\frac{2}{5})^{x - 1} = (\frac{25}{10})^{2(6x - 5)}\)

\((\frac{2}{5})^{x - 1} = (\frac{5}{2})^{12x - 10}\)

\((\frac{2}{5})^{x - 1} = (\frac{2}{5})^{-(12x - 10)}\)

When the bases are equal, the powers are also equal.

i.e. \(x - 1 = -12x + 10\)
\(13x = 11\)
\(x = \frac{11}{13}\)

Now,
Col. A: \((\frac{1}{x})^2\)

Col. B: \(2^x\)

Multipying both sides by \(x^2\);

Col. A: \(1\)
Col. B: \((x^2)(2^x)\)

Col. A: \(1\)
Col. B: \((\frac{11}{13})^2(2^{0.84})\)

Col. A: \(1\)
Col. B: \((\frac{121}{169})(2^{0.84})\)

Col. A: \(1\)
Col. B: \((0.7159)(2^{0.84})\)

We know, \(\sqrt{2} < 2^{0.84} < 2^1\), which means,

Col. A: \(1\)
Col. B: Greater than \(1\)

Hence, option B