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If 5^x = y, what is x? [#permalink]
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04 Nov 2014, 09:03
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Re: If 5^x = y, what is x? [#permalink]
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04 Nov 2014, 20:08
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5^x = y
We can therefore say, x = y^(1/5)
Statement 1. y^2 = 625 y can be +25 or 25
Since x,y should be real number y cannot be ve. Therefore, x= 25^(1/5). Sufficient
Statement 2. y^3=15625 therefore y = 25 and x = 25^(1/5). Sufficient.
Hence the answer is D)



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Re: If 5^x = y, what is x? [#permalink]
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04 Nov 2014, 20:38
kinghyts wrote: 5^x = y
We can therefore say, x = y^(1/5)
Statement 1. y^2 = 625 y can be +25 or 25
Since x,y should be real number y cannot be ve. Therefore, x= 25^(1/5). Sufficient
Statement 2. y^3=15625 therefore y = 25 and x = 25^(1/5). Sufficient.
Hence the answer is D) St1 says y^2=625.. Since LHS and RHS are greater than zero..We can take square root on both sides and get \(\sqrt{y^2}=\sqrt{625}\) or y=25 or y=+/25 So x can have 2 values... A is not sufficient.. If y=25 then x=2 and if y=25 then x=2 Ans is B as y^3=15625=positive...so y will have the same sign an y^3 so y is positive
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Re: If 5^x = y, what is x? [#permalink]
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04 Nov 2014, 22:05
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Theories: if \(a >= 0\) then\(a^x > 0\) ( *) So \(5^x > 0\), AND \(5^x = y\) > y must be positive. S1: \(y^2 = 625\) > \(y = 25\) OR \(y = 25\) From ( *), y must be positive > y is only 25 >\(5^x = 25 = 5^2\) > \(x = 2\) S1 is sufficient S2: \(y^3 = 15,625 = 25^3 > y = 25\) > \(5^x = 25 = 5^2\) > \(x = 2\) S2 is sufficient Ans: D
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Re: If 5^x = y, what is x? [#permalink]
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04 Nov 2014, 23:08
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WoundedTiger wrote: kinghyts wrote: 5^x = y
We can therefore say, x = y^(1/5)
Statement 1. y^2 = 625 y can be +25 or 25
Since x,y should be real number y cannot be ve. Therefore, x= 25^(1/5). Sufficient
Statement 2. y^3=15625 therefore y = 25 and x = 25^(1/5). Sufficient.
Hence the answer is D) St1 says y^2=625.. Since LHS and RHS are greater than zero..We can take square root on both sides and get \(\sqrt{y^2}=\sqrt{625}\) or y=25 or y=+/25 So x can have 2 values... A is not sufficient.. If y=25 then x=2 and if y=25 then x=2 Ans is B as y^3=15625=positive...so y will have the same sign an y^3 so y is positive 5^x = y , so if you take x=2 , it willbe 1/25 rt and we cant equate it with 25.... so y=25 then x=2 > how is it correct? Also, in 5^x = y we know LHS has positive base so RHS also should have positive base so 5^x can be equated only to 25, rt? Please clarify me.



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Re: If 5^x = y, what is x? [#permalink]
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04 Nov 2014, 23:39
anupamadw wrote: WoundedTiger wrote: kinghyts wrote: 5^x = y
We can therefore say, x = y^(1/5)
Statement 1. y^2 = 625 y can be +25 or 25
Since x,y should be real number y cannot be ve. Therefore, x= 25^(1/5). Sufficient
Statement 2. y^3=15625 therefore y = 25 and x = 25^(1/5). Sufficient.
Hence the answer is D) St1 says y^2=625.. Since LHS and RHS are greater than zero..We can take square root on both sides and get \(\sqrt{y^2}=\sqrt{625}\) or y=25 or y=+/25 So x can have 2 values... A is not sufficient.. If y=25 then x=2 and if y=25 then x=2 Ans is B as y^3=15625=positive...so y will have the same sign an y^3 so y is positive 5^x = y , so if you take x=2 , it willbe 1/25 rt and we cant equate it with 25.... so y=25 then x=2 > how is it correct? Also, in 5^x = y we know LHS has positive base so RHS also should have positive base so 5^x can be equated only to 25, rt? Please clarify me. Well...I goofed up on this one to be honest...Multitasking does this to you... Ans has to be D.. Thanks for pointing out
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Re: If 5^x = y, what is x? [#permalink]
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05 Nov 2014, 02:15
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I guess ans has to be D.
A  we get two options for y when y ^ 2 = 625 > +25 & 25. However, if we take a look at original statement 5 ^ x = y, only raising 5 to power of 4 will give 625. ve sign is maintained only for odd powers and therefore, even though 25 is one of the values, it is not valid. Hence A is suff.
B  y ^ 3 = 15625 > this says that y is +ve and we can obtain unique value for X from this option. Hence B is suff.
Ans D. My I missing anything here?
Ameya



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Re: If 5^x = y, what is x? [#permalink]
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Re: If 5^x = y, what is x?
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