Math Help - Solving for x

1. Solving for x

I'm supposed to solve for x,where x is a real number.

1. (x+3)(x-3)> 0 2. x^2 -2x-15 < 0

3. 12x^2=3x 4. sin2x = sinx, 0< x < 2pie

5. (x+1)^2(x-2) + (x+1)(x-2)^2=0 6. 27^2x=9^x-3

7. logx + log(x-3)=1

2. Originally Posted by journeygirl280
I'm supposed to solve for x,where x is a real number.

1. (x+3)(x-3)> 0
you begin by treating the inequality just as you would an equal sign. what would you do if you had (x + 3)(x - 3) = 0 ? well, if two numbers being multiplied gives zero, then one or the other must be zero, so we would split it up into two equations: x + 3 = 0 and x - 3 = 0. A similar thing is to be done here

(x + 3)(x - 3) > 0

=> x + 3 > 0 or x - 3 > 0

=> x > -3 or x > 3

now, with inequalities, things sometimes work out weird, and in evaluating algebraically, you might get the signs turned the wrong way. so here we test the answers. we have two values to work around, x = -3 and x = 3. x cannot be equal to either of these values since that would give zero, and we must be greater than zero. so now, let's check to the left and right of these numbers.

if x < - 3, is (x + 3)(x - 3) > 0 ? .........YES! so this is one solution

if -3 < x < 3, is (x + 3)(x - 3) > 0 ? .....NO! this isn't a solution

if x > 3, is (x + 3)(x - 3) > 0 ? .............YES! this is another solution

so finally, our solution is, x < -3 or x > 3

it looks long and tedious because i typed out the steps, but most of what i typed you should be able to do in your head somewhat easily, if not, very easily.

2. x^2 -2x-15 < 0
$x^2 - 2x - 15 \leq 0$

$\Rightarrow (x - 5)(x + 3) \leq 0$

Now continue the way I did in the previous question

3. 12x^2=3x
this is a nice one, there is no constant term to worry about, so we simply get everything on one side and factor out the common terms

$12x^2 = 3x$ ................put everything on one side

$\Rightarrow 12x^2 - 3x = 0$ ............now factor out the common term

$\Rightarrow 3x(4x - 1) = 0$

You finish this one. remember what i said about two numbers being multiplied give zero

3. Originally Posted by journeygirl280
4. sin2x = sinx, 0< x < 2pie
first of, it's "pi" not "pie"

We need to have a little background knowledge here, you should know that:

$\sin 2 \theta = 2 \sin \theta \cos \theta$

Now let's see how to do this:

$\sin 2x = \sin x$

$\Rightarrow 2 \sin x \cos x = \sin x$ .............applied the identity i mentioned above

$\Rightarrow 2 \sin x \cos x - \sin x = 0$ ..............got everything to one side

$\Rightarrow \sin x ( 2 \cos x - 1) = 0$

$\Rightarrow \sin x = 0$ or $2 \cos x - 1 = 0 \implies \cos x = \frac {1}{2}$

Now, what values of $x$ satisfy these equations, such that $0 \leq x \leq 2 \pi$ ?

5. (x+1)^2(x-2) + (x+1)(x-2)^2=0
Now this looks atrotous, but it's actually not that bad. with a keen eye you will realize that there are common terms here, they don't look like regular terms, like say, x, but they are groups. we have (x + 1) and (x - 2) being common to both terms here, so let's factor them out and see where it gets us

$(x + 1)^2 (x - 2) + (x + 1)(x - 2)^2 = 0$ ...........factor out the common terms

$\Rightarrow (x + 1)(x - 2)[(x + 1) + (x - 2)] = 0$ .........simplify what's in the square brackets

$\Rightarrow (x + 1)(x - 2)(2x - 1) = 0$

Can you take it from here?

6. 27^2x=9^x-3
Please use parentheses. it seems that (x - 3) is the power of 9, but i'm not sure. be careful when you are typing math, brackets are important tools to clarify what you mean

i'll assume x - 3 is the power of 9 (which means you should have typed 27^(2x) = 9^(x - 3) )

I will further assume that you are familiar with the laws of exponents, if not, look them up.

$27^{2x} = 9^{x - 3}$

this is a nice exponential equation, we can express both side in terms of the same base. Note that 27 = 3^3 and 9 = 3^2, so we can replace 27 and 9 with their representations in base 3

$\Rightarrow \left( { \color {red} 3^3 } \right)^{2x} = \left( { \color {red} 3^2 } \right)^{x - 3}$

Now we multiply the powers (i hope you know why, here's where your knowledge of the laws of exponents comes into play)

$\Rightarrow 3^{6x} = 3^{2x - 6}$

Now if the bases are the same, the powers have to be the same to maintain equality, so we can equate the powers.

$\Rightarrow 6x = 2x - 6$

the rest is trivial, i leave it to you

7. logx + log(x-3)=1
Ah, logarithms. I hope you know the rules. here are the ones we need for this problem:

Law 1:
If $\log_a b = c$, then $a^c = b$

Law 2: $\log_a x + \log_a y = \log_a xy$

Now, on to the problem (I assume we are using log to the base 10, as is reasonable):

$\log x + \log (x - 3) = 1$

$\Rightarrow \log x(x - 3) = 1$ ...........applied law 2

$\Rightarrow 10^1 = x(x - 3)$ .............applied law 1

$\Rightarrow x^2 - 3x - 10 = 0$

I think you can take it from here

Godspeed and goodluck!