If 9x=27^y, which of the following expresses x in terms of y

9x = 27^y
x = 27^y/9
= 3^3y/3^2
= 3^(3y-2)

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

Answer: E.

For theory on exponents check: math-number-theory-88376.html

Tough DS questions on exponents with solutions: tough-and-tricky-exponents-and-roots-questions-125967.html
Tough PS questions on exponents with solutions: tough-and-tricky-exponents-and-roots-questions-125956.html

Hope it helps.

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.

9x=27^y

3^2x=3^3y

devide both sides by 3^2

x=3^3y-2.

note, you need to make both sides have equale exponent, and remember when you simplify exponents through division, you subtract them

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

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.

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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9x=27^y
3^2 x = 3^3y

Clearly, x=1 an y=2/3 will satisfy

Now we have the values of x given in the options, substitute y=2/3 , wherever we get value =1, is the answer

which is clearly E since only that has 3 multiplied with y which conveniently cancels our denominator and results in 3^0 =1

Ans=>E

27 can be written as 9^ 3/2

So x = 9^(3y/2-1)

x = 3^(3y-2) = Answer = E

\((3)^{2}\)x= \((3)^{3y}\)

So x= \((3)^{3y}\)/\((3)^{2}\)

x= \((3)^{3y-2}\)

Answer: E.

Kudos please if you like my explanation!

Let’s simplify the given equation:

9x = 27^y

(3^2)x = 3^(3y)

x = 3^(3y)/3^2

x = 3^(3y-2)

Answer: E

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