If y^c = y^d+1 , what is the value of y?

Hi All,

Here, we're told that Y^C = Y^(D+1). We're asked for the value of Y.

Since the 'base' of each expression is the same, you might be tempted to think that C = D+1. That will be true in MOST situations, but NOT necessarily true in others (e.g. when Y = -1, 0 or 1). We'll have to consider more than just the 'obvious' values of Y to get this question correct.

Fact 1: Y < 1

The exponents have NO impact on the value of Y, so Y can be ANY value that is less than 1.

Y could be 0
Y could be -1
Etc.
Fact 1 is INSUFFICIENT

Fact 2: D = C

IF...
D=C=1
Y^1 = Y^2
Y = 0 or 1
Fact 2 is INSUFFICIENT

Combined, we know....
Y < 1
D=C

From our prior work, we know that Y could be 0, but we have to do a bit of work to prove that it can't be anything else....

IF...
D=C=0
Y^0 = Y^1
Y = 1, but we know from Fact 1 that Y must be LESS than 1, so this answer is not an option.

IF...
D=C=2
Y^2 = Y^3
Y = 0 or 1; from our prior work, we know that 0 is an option and 1 is NOT.

IF...
D=C=-1
Y^(-1) = Y^0
Y = 1; again, not an option

IF....
D=C=1/2
Y^(1/2) = Y^(3/2)
Y = 0 or 1
Etc.

From this, the only options that occur are 0 and 1 (which is not an allowed). Thus, Y ALWAYS is 0.
Combined, SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

A is the answer..
Statement 1 alone is sufficient I think...


>>>
We can infer from the question stem that y can take only 2 values. 0 or 1.

Statement says that y < 1.

Hence y can only be 0. Nothing else.


if y=-1, then the statement Y^C = Y^(D+1) is never true.



Am I ignoring something?
I am doubting on the OA.

What if y = -1 and both c and d+1 are even? In this case (-1)^(even) = (-1)^(even) = 1.
Or what if c = d+1? In this case y can take any value.

Hi,
can I get a simpler explanation or some official explanation of this problem. as i still couldn't understand the same.
Thanks
Celestial

Hi Celestial09,

The question tests the concept of special cases of exponents where the final value of the expression \(base^{exponent}\) is independent of the value of exponent. There are three such cases possible:

1. If \(base = 0\) , then \(base^{exponent}= 0\) irrespective of the value of the exponent as \(0\) raised to any power is \(0\)

2. If \(base = 1\), then \(base^{exponent} = 1\) irrespective of the value of the exponent as \(1\) raised to any power is \(1\)

3. If \(base = -1\), then \(base ^{Even exponent} = 1\) & \(base^{Odd exponent} = -1\), irrespective of any even or odd number in the exponent

For all the other values of bases if \(base^{exponent1} = base^{exponent2}\), we can equate the values of the exponents. We are asked to find the value of y(which is the base in this case). Let's see if the statements allow us to find a unique value of y.

Statement- I:
St-I tells us that \(y < 1\).

Thus \(y\) can take a value of \(-1\) or \(0\) from our special cases or any other value when the exponents are equal. Clearly insufficient to get a unique value of \(y\).

Statement-II:
St-II tells us that d=c.

This statement does help us narrow down a few values of \(y\). Let's consider the general cases first.
If \(y\) is any value we can equate the exponents i.e. c= d +1. Since c=d, we can't write c = d+1. This means that the expression \(base^{exponent}\) is independent of the base(which we have defined as our special cases)

From the special cases of \(y = {-1, 0, 1}\) we can eliminate -1 because if c is even then d+1 would be odd and vice versa. Thus we will have different odd & even exponents on both sides of the equations.
However, we still can't say for sure if \(y\) =\(0\) or \(1\) as in both the cases, c can be equal to d.

Thus, statement-II also is insufficient to answer our question.

Combination of Statement-I & II:
St-II tells us that \(y = {0, 1}\) and st-I tells us that \(y < 1\).

Combining these two analysis, we can eliminate \(y = 1\) as our answer which leaves us with only one value of \(y = 0\).

Thus combination of the statements is sufficient to answer our question.

Hope its clear! Let me know if you are having trouble at any point of this solution.

Regards
Harsh

A few considerations:

case 1) If y = 1, then y^c = y^(d+1) for ALL values of c and d
case 2) If y = 0, then y^c = y^(d+1) for ALL values of c and d
case 3) If y = -1, then y^c = y^(d+1) if c and (d+1) are both ODD or both EVEN
case 4) If y equals any value OTHER THAN 1, 0 or -1, then y^c = y^(d+1) only if c and (d+1) are EQUAL

Target question: What is the value of y?

Given: y^c = y^(d+1)

Statement 1: y < 1
This rules out case 1 above.
However, this still leaves several possibilities that yield conflicting answers to the target question.
Here are two:
Case a: y = 0, c = 1 and d = 1. Notice that this satisfies the given information [y^c = y^(d+1)] because 0^1 = 0^(1+1). In this case y = 0
Case b: y = 3, c = 2 and d = 1. Notice that this satisfies the given information [y^c = y^(d+1)] because 3^2 = 3^(1+1). In this case y = 3
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: d = c
This rules out cases 3 and 4 above. Here's why:
case 3: if c and d are equal, then c and (d+1) CANNOT both be EVEN or ODD
case 4: if c and d are equal, then c and d+1 cannot be equal
However, this still leaves several possibilities that yield conflicting answers to the target question.
Here are two:
Case a: y = 0, c = 1 and d = 1. Notice that this satisfies the given information [y^c = y^(d+1)] because 0^1 = 0^(1+1). In this case y = 0
Case b: y = 1, c = 1 and d = 1. Notice that this satisfies the given information [y^c = y^(d+1)] because 1^1 = 1^(1+1). In this case y = 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 rules out case 1 above
Statement 2 rules out cases 3 and 4 above
This leaves only case 2, which means y must equal 0
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer:

Cheers,
Brent

\(y^{c} = y^{d+1}\)

We have 3 variables here: y, c and d
We can easily make LHS = RHS by different methods:
1. c = d + 1 OR
2. y = 0 or 1 or in some cases -1 (when both c and d + 1 are even)

1) y<1
y can take many values such as 1/2, 0, -1 or -5 etc. All we need to do is make c = d + 1 such as c = 3 and d = 2 to make LHS = RHS.
Not sufficient.

2) d=c
This tells us that c is not equal to d + 1 since it is equal to d. So y could be 0 or 1, not -1 (because c and d+1 are consecutive values so they both cannot be even).
Not sufficient.

Using both statements,
Stmnt 2 tells us that y is 0 or 1 but stmnt 1 tells us that y is not 1. So y MUST BE 0.
Sufficient.

Answer (C)

If \(y^c = y^{(d + 1)}\), what is the value of y ?

(1) y < 1
(2) d = c

From the main question stem we can say c=d+1 which means that c & d are consecutive numbers..

frm 1; y<1, it does not specify anything about c& d and y can be any value less than 1 so insufficient..

Frm 2: d=c

let c=d=1

using the values the eqn would become
y^2=y^3
1=y
and for any integer value of c& d we would get y=1 but in case of any fraction value of c and d not one answer so insufficient..

combining 1& 2
from given eqn we would always get y= 1
IMO C ...

hi archit if y=1 then doesn't it contradict the 1st statement? i think the value should be 0 as y<1 in the first statement.

KaulMeAnkit
Hello,

See y can be any value <1 , i.e 1/2, 1/3 or 0 , it specifically cannot be '0'

what I actually meant by :
"combining 1& 2
from given eqn we would always get y= 1"

is that upon combining both 1& 2 , for any value of y <1 , we would always get a y=1 for eg if y=1/2 and c =d=1
then
(1/2)^1=(1/2)^2
or
(1/2)=1 or say y=1 ' y is 1/2'

Beautiful conceptual problem, Bunuel. Congrats! (kudos!)

Important : \(\,\,{y^c} = {y^{d + 1}}\,\,\,\, \Rightarrow \,\,\,c = d + 1\,\,\,\,\,\,\underline {{\rm{if}}} \,\,\,\,y \notin \left\{ { - 1,0,1} \right\}\,\,\,\,\left( * \right)\)

\(? = y\)

\(\left( 1 \right)\,\,y < 1\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {y,c,d} \right) = \left( {0,1,1} \right)\,\,\,\, \Rightarrow \,\,? = 0\,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {y,c,d} \right) = \left( { - 1,1,0} \right)\,\,\,\, \Rightarrow \,\,\,? = - 1\,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,d = c\,\,\left\{ \matrix{\\
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {y,c,d} \right) = \left( {0,1,1} \right)\,\,\,\, \Rightarrow \,\,? = 0\,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {y,c,d} \right) = \left( {1,1,1} \right)\,\,\,\, \Rightarrow \,\,\,? = 1\,\, \hfill \cr} \right.\,\)

\(\left. {\left( {1 + 2} \right)\,\,\,\left\{ \matrix{\\
\,y = 0\,\,{\rm{with}}\,\,c = d\,\,\,{\rm{viable}}\,\,\,\left( {{\rm{i}}{\rm{.e}}{\rm{.,}}\,\,{\rm{not}}\,\,{\rm{creating}}\,\,{0^0}} \right)\,\,\, \Rightarrow \,\,\,\,\,? = 0 \hfill \cr \\
\,y = - 1\,\,\,\, \Rightarrow \,\,\,\,\,{\left( { - 1} \right)^{c - \left( {d + 1} \right)}} = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,c - \left( {d + 1} \right) = {\rm{even}}\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\,\,\,\,\left( {d = c} \right)\,\,\,\,\, \hfill \cr \\
\,y < 1\,\,,\,\,\,y \ne 0\,\,,\,\,y \ne - 1\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)\,} \,\,\,\,\,c = d + 1\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\,\,\,\,\left( {d = c} \right)\, \hfill \cr} \right.\,\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = 0\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.