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What's wrong if i solve the problem as follows: 3*3x=3*9^y 3x = 9^y [Dividing both side by 3] x = 3^(2y-1) Ans. C. Please help to correct my wrong concept as the OA is E. _________________

What's wrong if i solve the problem as follows: 3*3x=3*9^y 3x = 9^y [Dividing both side by 3] x = 3^(2y-1) Ans. C. Please help to correct my wrong concept as the OA is E.

The point is that \(3*9^y\) does not equal to \(27^y\): \(3*9^y=3*3^{2y}=3^{2y+1}\) on the other hand \(27^y=(3^3)^y=3^{3y}\).

Solution: If 9x = 27^y, which of the following expresses x in terms of y? A. 3^y B. 3^(y-1) C. 3^(2y-1) D. 3^(2y-3) E. 3^(3y-2)

\(9x=3^2*x\) and \(27^y=(3^3)^y=3^{3y}\) --> \(3^2*x=3^{3y}\) --> \(x=\frac{3^{3y}}{3^2}\) --> \(x=3^{3y-2}\).

What's wrong if i solve the problem as follows: 3*3x=3*9^y 3x = 9^y [Dividing both side by 3] x = 3^(2y-1) Ans. C. Please help to correct my wrong concept as the OA is E.

Bunuel has already pointed out your error so I will not repeat it.

Let me give you another method of working out the solution here (though you should ensure that you understand the theory of exponents well)

If 9x = 27^y, which of the following expresses x in terms of y? A. 3^y B. 3^(y-1) C. 3^(2y-1) D. 3^(2y-3) E. 3^(3y-2)

We need x in terms of y. In 9x = (27)^y, put y = 0. You get x = 1/9

Now see which option will give you 1/9 when you put y = 0. I hope you can quickly see that only option (E) will give you x = 1/9 when y = 0. Answer is (E). Such methods work well when you have variables in the options.

Note: If more than one options had given 1/9, you could have tried some other values to choose the right answer out of those options. _________________

Re: If 9x=27^y, which of the following expresses x in terms of y [#permalink]

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08 Apr 2012, 22:41

To Karishma,

I'm glad you used the same approach as I've used, but I used y=1, which makes, 9x = 27^y 9x = 27 x = 3. Look for options which give ans. as x=3, I marked A (option E also gives x=3), which gives x=3 which is wrong ans. Can u explain where I'm wrong

I'm glad you used the same approach as I've used, but I used y=1, which makes, 9x = 27^y 9x = 27 x = 3. Look for options which give ans. as x=3, I marked A (option E also gives x=3), which gives x=3 which is wrong ans. Can u explain where I'm wrong

Thnx, Priyal

When you put y = 1, three of the five options give you x = 3 (options A, C and E). Any one of these 3 could be the correct answer. You now need to try out some other values of y to get the answer out of these 3 options.

You need to check all the options to ensure that no other options gives you the same value. When I put y = 0, I try it in all the options and only option E gives me 1/9 so I can directly mark that as the answer. This is the reason I mentioned the note in the post above: "Note: If more than one options had given 1/9, you could have tried some other values to choose the right answer out of those options."

So now your next step is to put y = 0 and out of the 3 options, only E will satisfy x = 1/9. Try and put the easiest value first. Easiest value is generally 0, if allowed.

Putting y = 2/3/4... will make it cumbersome. _________________

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