1. ## Find

1.
Find: $\displaystyle \sqrt{15-x}+\sqrt{4-x}= ?, if \sqrt{15-x}-\sqrt{4-x}=2$
2.
$\displaystyle a^b$ = 81, $\displaystyle b^c$ = 2, $\displaystyle a^c$ = 3. Find $\displaystyle b^b$

2. Originally Posted by Ellla
1.
Find: $\displaystyle \sqrt{15-x}+\sqrt{4-x}= ?, if \sqrt{15-x}-\sqrt{4-x}=2$
2.
$\displaystyle a^b$ = 81, $\displaystyle b^c$ = 2, $\displaystyle a^c$ = 3. Find $\displaystyle b^b$
It's not clear what you are trying to do for number one. Are you trying to solve for $\displaystyle x$?

Secondly, you should show that you've attempted to solve the problem. We're not suppose to just give answers, since people will obviously abuse the forum. However, it's understandable if you don't know how to get started on a problem.

When you need to solve a radical equation, isolate one of the radicals and square.

$\displaystyle \sqrt{15-x}-\sqrt{4-x}=2$

$\displaystyle \sqrt{15-x}=2+\sqrt{4-x}$

$\displaystyle 15-x=4+2\sqrt{4-x}+(4-x)$

Now isolate the other radical and square again. Then solve for $\displaystyle x$. See if you can figure out how to do that. If you can't, just repsond with your attempt and we will help.

3. Originally Posted by Ellla
1.
Find: $\displaystyle \sqrt{15-x}+\sqrt{4-x}= ?, if \sqrt{15-x}-\sqrt{4-x}=2$
Use "$\displaystyle (a-b)(a+b)= a^2- b^2$"
$\displaystyle \left(\sqrt{15-x}- \sqrt{4- x}\right)\left(\sqrt{15-x}+ \sqrt{4-x}\right)= \left(\sqrt{15-x)\right)^2- \left(\sqrt{4-x}\right)^2$[math
2.
$\displaystyle a^b$ = 81, $\displaystyle b^c$ = 2, $\displaystyle a^c$ = 3. Find $\displaystyle b^b$
There should be a "direct" way but this will work: take the logarithm, base 3, of each of the equations.

4. Your second problem is kind of confusing, but I think I've solved it using logarithms. There may be a simple solution, but here's what I have:

First note that the equation $\displaystyle a^b=81$ means that:
$\displaystyle b=log_a(81)=log_a(9^2)=2log_a(9)=4log_a(3)$

So $\displaystyle b=4log_a(3)$.

-----------------------------------------------------
The equations $\displaystyle b^c=2$ and $\displaystyle a^c=3$ give $\displaystyle c=log_b(2)$ and $\displaystyle c=log_a(3)$. This means that:

$\displaystyle log_b(2)=log_a(3)$

$\displaystyle 4log_b(2)=4log_a(3)$

$\displaystyle 4log_b(2)=b$

$\displaystyle log_b(2^4)=b$

$\displaystyle log_b(16)=b$

This means that $\displaystyle b^b=16$.

5. hi
$\displaystyle \left (\sqrt{15-x}+\sqrt{4-x} \right )\left ( \sqrt{15-x}-\sqrt{4-x} \right )=2y$
find $\displaystyle y$.

6. well for the first one cant you square both sides???
15-x-4-x = 4
-2x = -7
x=7/2

or am i wrong?

7. Originally Posted by Raoh
hi
$\displaystyle \left (\sqrt{15-x}+\sqrt{4-x} \right )\left ( \sqrt{15-x}-\sqrt{4-x} \right )=2y$
find $\displaystyle y$.
remember: $\displaystyle a^2 + b^2 = (a - b) (a + b)$

8. Originally Posted by pencil09
remember: $\displaystyle a^2 + b^2 = (a - b) (a + b)$
$\displaystyle (a-b)(a+b)=a^{2}-b^{2}$

9. let $\displaystyle \sqrt{15 - x} + \sqrt{4 - x} = y$, then
$\displaystyle (\sqrt{15 - x} + \sqrt{4 - x})(\sqrt{15 - x} - \sqrt{4 - x}) = (y)(2) = 2y$
hence $\displaystyle (\sqrt{15 - x})^2 - (\sqrt{4 - x})^2 = 2y$
15 - x - 4 + x = 2y
2y = 11
y = $\displaystyle \frac{11}{2}$

10. Originally Posted by Raoh
$\displaystyle (a-b)(a+b)=a^{2}-b^{2}$
ha,,ha,,, sorry my mistake...

11. Hello, Ellla!

$\displaystyle \begin{array}{cccc}a^b &=&81 & {\color{blue}[1]} \\ b^c &=& 2 & {\color{blue}[2]} \\ a^c &=& 3 & {\color{blue}[3]}\end{array}$

$\displaystyle \text{Find }\,b^b$

$\displaystyle \text{From }{\color{blue}[1]}\!:\;\;a \:=\:81^{\frac{1}{b}} \:=\:(3^4)^{\frac{1}{b}} \quad\Rightarrow\quad a\:=\:3^{\frac{4}{b}}\;\;{\color{blue}[4]}$

$\displaystyle \text{From }{\color{blue}[3]}\!:\;\;a \:=\:3^{\frac{1}{c}}\;\;{\color{blue}[5]}$

$\displaystyle \text{Equate }{\color{blue}[4]}\text{ and }{\color{blue}[5]}\!:\;\;3^{\frac{4}{b}} \:=\:3^{\frac{1}{c}} \quad\Rightarrow\quad \frac{4}{b} \:=\:\frac{1}{c} \quad\Rightarrow\quad c \:=\:\frac{b}{4}$

$\displaystyle \text{Then }{\color{blue}[2]}\text{ becomes: }\;b^{\frac{b}{4}} \:=\:2$

$\displaystyle \text{Raise to the 4th power: }\;\left(b^{\frac{b}{4}}\right)^4 \:=\:2^4 \quad\Rightarrow\quad b^b \:=\:16$

12. Hello again, Ellla!

2. Given: .$\displaystyle \sqrt{15-x}-\sqrt{4-x}=2$

Find $\displaystyle \sqrt{15-x} + \sqrt{4-x}.$
Multiply both sides by: .$\displaystyle (\sqrt{15-x} + \sqrt{4-x})$

. . $\displaystyle (\sqrt{15-x} - \sqrt{4-x})(\sqrt{15-x} + \sqrt{4-x}) \;=\;2(\sqrt{15-x} + \sqrt{4-x})$

. . . . . . . . . . . . . . . .$\displaystyle (15-x) - (4-x) \;=\;2(\sqrt{15-x} + \sqrt{4-x})$

. . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle 11 \:=\:2(\sqrt{15-x} + \sqrt{4-x})$

$\displaystyle \text{Therefore: }\;\sqrt{15-x} + \sqrt{4-x} \;=\;\frac{11}{2}$