problems with logs and a proof.

**driver327** · June 26th 2007, 05:59 PM

1. I used the intermediate value theorem for (square root x to the third)= x^2+2x-1 on (0,1), I plugged 0 and 1, and I got (left side)0=-1(right side) and 1=2 respectively. Since I got a negative(on 0), and positive on the other(on 1), there is at least one solution for (0,1), is this right? If I have the wrong idea, tell me. 2. 1/2 lnx/ln3=ln8/ln2. ans. 729. 3. (square root to the third 2)^x + 10=2^(x^2). ans. 2, -5/3 4.(ln x/ln 3)+(ln 2x+1/ln 3)=1 ans. 1 I do not know how to put the small numbers on logs so I put it in natural log form. Thanks in advance.

**topsquark** · June 26th 2007, 06:14 PM

Originally Posted by driver327

1. I used the intermediate value theorem for (square root x to the third)= x^2+2x-1 on (0,1), I plugged 0 and 1, and I got (left side)0=-1(right side) and 1=2 respectively. Since I got a negative(on 0), and positive on the other(on 1), there is at least one solution for (0,1), is this right? If I have the wrong idea, tell me.

Wow. Ummmm... Let me break this down a bit. You've got the function:
$\sqrt{x^3} = x^2 + 2x - 1$
on the interval (0, 1).

Are you trying to find a solution for x?

If so then try moving everything to the left:
$\sqrt{x^3} - (x^2 + 2x - 1) = 0$

NOW plug in x = 0 and x = 1:
$\sqrt{0^3} - (0^2 + 2 \cdot 0 - 1) = 1$

$\sqrt{1^3} - (1^2 + 2 \cdot 1 - 1) = -1$

Because the first is positive and the second negative we know that it (the LHS) passes through 0 at some point in the interval.

-Dan

**topsquark** · June 26th 2007, 06:17 PM

Originally Posted by driver327

2. 1/2 lnx/ln3=ln8/ln2. ans. 729.

Again, solve for x?

$\frac{1}{2} \frac{ln(x)}{ln(3)} = \frac{ln(8)}{ln(2)}$

Now,
$ln(8) = ln(2^3) = 3 \cdot ln(2)$ , so

$\frac{1}{2} \frac{ln(x)}{ln(3)} = \frac{3 \cdot ln(2)}{ln(2)}$

$\frac{1}{2} \frac{ln(x)}{ln(3)} = 3$

$\frac{ln(x)}{ln(3)} = 6$

$ln(x) = 6 \cdot ln(3)$

$ln(x) = ln(3^6) = ln(729)$

$x = 729$

-Dan

**topsquark** · June 26th 2007, 06:20 PM

Originally Posted by driver327

3. (square root to the third 2)^x + 10=2^(x^2). ans. 2, -5/3

Is this $(\sqrt{2}~^3)^x + 10 = 2^{x^2}$ ?

-Dan

**topsquark** · June 26th 2007, 06:24 PM

Originally Posted by driver327

4.(ln x/ln 3)+(ln 2x+1/ln 3)=1 ans. 1

Is this supposed to be
$\frac{ln(x)}{ln(3)} + \frac{ln(2x + 1)}{ln(3)} = 1$ ?

$ln(x) + ln(2x + 1) = ln(3)$

$ln(x(2x + 1)) = ln(3)$

$x(2x + 1) = 3$

$2x^2 + x - 3 = 0$

$(2x + 3)(x - 1) = 0$

So
$x = -\frac{3}{2}$
or
$x = 1$

But a negative x is outside the domain of the expression on the LHS. Thus the only solution is x = 1.

-Dan

**driver327** · June 26th 2007, 10:04 PM

Originally Posted by topsquark

Is this $(\sqrt{2}~^3)^x + 10 = 2^{x^2}$ ?

-Dan

The 3 was on top of the square root, and the +10 was part of the exponent. Sorry for the confusion.

**topsquark** · June 26th 2007, 10:20 PM

Originally Posted by driver327

The 3 was on top of the square root, and the +10 was part of the exponent. Sorry for the confusion.

So it's
$\sqrt[3]{2^{(x+10)}} = 2^{x^2}$ <-- This I can work with!

Well, $\sqrt[3]{x} = x^{1/3}$ so:
$(2^{(x+10)})^{1/3} = 2^{x^2}$

$(2^{(x+10)/3}) = 2^{x^2}$

Thus
$\frac{x + 10}{3} = x^2$

$x + 10 = 3x^2$

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

$(3x + 5)(x - 2) = 0$

Thus
$x = -\frac{5}{3}$ or $x = 2$ .

Both solutions are in the domain of the original expression, so both are solutions.

-Dan

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