I'm so confused on these two problems. If you could help at all I would really appreciate it. Thanks!
Hi there, in the first question, are they addition signs or division signs underneath the square root?
In the second quesiton use some logs.
$\displaystyle 2^x = 15 \Rightarrow x = \log_215$
$\displaystyle 15^y = 32 \Rightarrow y = \log_{15}32$
$\displaystyle x\times y = \log_215 \times \log_{15}32=\dots$
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.
$\displaystyle \sqrt{n^2 + n^2 + n^2 + n^2} = 64$
$\displaystyle \sqrt{4n^2} = 64$
$\displaystyle 2 \sqrt{n^2} = 64$
$\displaystyle \sqrt{n^2} = 32$
Since $\displaystyle n > 0$, $\displaystyle \sqrt{n^2} = n$
So, $\displaystyle n = 32$.
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$\displaystyle 2^x = 15$, 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, $\displaystyle xy = \frac{ln(15)}{ln(2)} \times \frac{ln(32)}{ln(15)}$
$\displaystyle xy = \frac{ln(15) \times ln(32)}{ln(2) \times ln(15)}$
$\displaystyle xy = \frac{ln(32)}{ln(2)} = 5$