IanStewart wrote:
Bunuel wrote:
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{5^{(x-1)^5}}{4^{-\sqrt{y}}\)?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14
5. If \(x=23^2*25^4*27^6*29^8\) and is a multiple of \(26^n\), where \(n\) is a non-negative integer, then what is the value of \(n^26-26^n\)?
A. -26
B. -25
C. -1
D. 0
E. 1
Welcome back Bunuel!
Some nice questions. A couple of comments:
Q5 has a typo at the end: \(n^26-26^n\). I think you need to include the exponent 26 in braces {26} to get \(n^{26}-26^n\)
In Q3, unless I'm making some kind of stupid mistake, the value of the expression in the question is close to 600 (so isn't among the answers). I'm wondering if you meant the question to instead read something more like:
3. If \(5^{10x}=4,900\) and \(4^{\sqrt{y}}=25\) what is the value of \(\frac{5^{5x-1}}{2^{\sqrt{y}}\)?A. 14/5
B. 5
C. 28/5
D. 13
E. 14
In that case, I get one of the five answer choices (A).
The answer should be E. 14
Here is how:
\(5^{10x}=4900\)
so \({(5^{5x})}^2=70^2\)
Take square root of both sides >
so \(5^{5x}=70\)
Also from question stem:
\(2^{\sqrt{y}} = 5^2\)
So (from the question stem)
\(4^{-\sqrt{y}} = \frac{1}{4^{\sqrt{y}}}\)
\(4^{-\sqrt{y}} = (\frac{1}{2^{\sqrt{y}}})^2 = \frac{1}{5^4}\)
Now \(5^{(x-1)^5}\)
Should be \(\frac{5^{5x}}{5^5}\)
We know \(5^{10x}=4900\)
so \(5^{10x}=70^2\)
so \(5^{5x}=70\)
So \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)
is basically \(\frac{5^x}{5^5}*5^4\)
We already know the value of \(5^x\) which is \(70\)
So now it becomes \(\frac{70}{5^5}*5^4\)
Which should resolve to \(14\)
Hence Answer = E = \(14\)