Find

**Ellla** · November 8th 2009, 05:10 AM

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

**adkinsjr** · November 8th 2009, 05:45 AM

Originally Posted by Ellla

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

It's not clear what you are trying to do for number one. Are you trying to solve for $x$ ?

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.

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

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

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

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

**HallsofIvy** · November 8th 2009, 06:15 AM

Originally Posted by Ellla

1.
Find: $\sqrt{15-x}+\sqrt{4-x}= ?, if \sqrt{15-x}-\sqrt{4-x}=2$

Use " $(a-b)(a+b)= a^2- b^2$ "
$\left(\sqrt{15-x}- \sqrt{4- x}\right)\left(\sqrt{15-x}+ \sqrt{4-x}\right)= \left(\sqrt{15-x)\right)^2- \left(\sqrt{4-x}\right)^2$ [math

2.
$a^b$ = 81, $b^c$ = 2, $a^c$ = 3. Find $b^b$

There should be a "direct" way but this will work: take the logarithm, base 3, of each of the equations.

**adkinsjr** · November 8th 2009, 06:15 AM

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 $a^b=81$ means that:
$b=log_a(81)=log_a(9^2)=2log_a(9)=4log_a(3)$

So $b=4log_a(3)$ .

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

$log_b(2)=log_a(3)$

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

$4log_b(2)=b$

$log_b(2^4)=b$

$log_b(16)=b$

This means that $b^b=16$ .

**Raoh** · November 8th 2009, 06:40 AM

hi

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

**orange gold** · November 8th 2009, 06:49 AM

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

or am i wrong?

**pencil09** · November 8th 2009, 08:16 AM

Originally Posted by Raoh

hi

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

remember: $a^2 + b^2 = (a - b) (a + b)$

**Raoh** · November 8th 2009, 08:24 AM

Originally Posted by pencil09

remember: $a^2 + b^2 = (a - b) (a + b)$

$(a-b)(a+b)=a^{2}-b^{2}$

**ukorov** · November 8th 2009, 08:33 AM

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

**pencil09** · November 8th 2009, 05:06 PM

Originally Posted by Raoh

$(a-b)(a+b)=a^{2}-b^{2}$

ha,,ha,,, sorry my mistake...

**Soroban** · November 8th 2009, 06:43 PM

Hello, Ellla!

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

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

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

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

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

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

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

**Soroban** · November 8th 2009, 06:55 PM

Hello again, Ellla!

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

Find $\sqrt{15-x} + \sqrt{4-x}.$

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

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

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

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

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

**istenum** · November 8th 2009, 07:28 PM

bookmarked and b back l8er, bro, :-)
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