# Two problems- one with square root, one with powers.

• Dec 13th 2009, 12:00 PM
lucyluu
Two problems- one with square root, one with powers.
http://i49.tinypic.com/28guur5.jpg

I'm so confused on these two problems. If you could help at all I would really appreciate it. Thanks!
• Dec 13th 2009, 12:19 PM
Raoh
Quote:

Originally Posted by lucyluu
http://i49.tinypic.com/28guur5.jpg

I'm so confused on these two problems. If you could help at all I would really appreciate it. Thanks!

but anyway,
$\displaystyle \sqrt{n^2+n^2+n^2+n^2}$$\displaystyle =\sqrt{4n^2}=\sqrt{(2n)^2}$$\displaystyle =\left | 2n \right |=2n=64$
therefore $\displaystyle n=32$
• Dec 13th 2009, 12:20 PM
pickslides
Hi there, in the first question, are they addition signs or division signs underneath the square root?

In the second quesiton use some logs.

$\displaystyle 2^x = 15 \Rightarrow x = \log_215$

$\displaystyle 15^y = 32 \Rightarrow y = \log_{15}32$

$\displaystyle x\times y = \log_215 \times \log_{15}32=\dots$
• Dec 13th 2009, 12:22 PM
Raoh
Quote:

Originally Posted by pickslides
Hi there, in the first question, are they addition signs or division signs underneath the square root?

if they were division signs,there would be no longer $\displaystyle n$ to find(Rofl)
• Dec 13th 2009, 12:37 PM
lucyluu
Quote:

Originally Posted by pickslides
Hi there, in the first question, are they addition signs or division signs underneath the square root?

In the second quesiton use some logs.

$\displaystyle 2^x = 15 \Rightarrow x = \log_215$

$\displaystyle 15^y = 32 \Rightarrow y = \log_{15}32$

$\displaystyle x\times y = \log_215 \times \log_{15}32=\dots$

Thank you for the response. However, my teacher hasn't taught us anything about logs yet... is there any other way to do it?
• Dec 13th 2009, 12:45 PM
Defunkt
$\displaystyle 2^x = 15, \ 15^y = 32 \Rightarrow 15^y = (2^x)^y = 32 = 2^5$
• Dec 13th 2009, 12:51 PM
pickslides
You can find some estimates by trial and error.

Consider $\displaystyle 2^x = 15$

We know that $\displaystyle 2^3 = 8$ and $\displaystyle 2^4 = 16$ as 15 is between 8 and 16 then x must be between 3 and 4, probably closer to 4.

It now depends on how accuarate you want to be.

Do the same for $\displaystyle 15^y = 32$, the answer should be between 1 and 2.
• Dec 13th 2009, 12:54 PM
Raoh
it seems that he only wanted the value of $\displaystyle xy$ which is $\displaystyle 5$ (as "Defunkt" showed).
• Dec 13th 2009, 01:04 PM
Bacterius
$\displaystyle \sqrt{n^2 + n^2 + n^2 + n^2} = 64$

$\displaystyle \sqrt{4n^2} = 64$

$\displaystyle 2 \sqrt{n^2} = 64$

$\displaystyle \sqrt{n^2} = 32$

Since $\displaystyle n > 0$, $\displaystyle \sqrt{n^2} = n$

So, $\displaystyle n = 32$.

--------------------------

$\displaystyle 2^x = 15$, so $\displaystyle x = log_2(15)$

$\displaystyle 15^y = 32$, so $\displaystyle y = log_{15}(32)$

Thus $\displaystyle xy = log_2(15) \times log_{15}(32)$.

Now, you know from the change of base formula that :

$\displaystyle log_2(15) = \frac{ln(15)}{ln(2)}$

$\displaystyle log_{15}(32) = \frac{ln(32)}{ln(15)}$

So, $\displaystyle xy = \frac{ln(15)}{ln(2)} \times \frac{ln(32)}{ln(15)}$

$\displaystyle xy = \frac{ln(15) \times ln(32)}{ln(2) \times ln(15)}$

$\displaystyle xy = \frac{ln(32)}{ln(2)} = 5$

:)