1. ## Random Equation Help?

Obviously, a calculator cannot be used.

1)
Code:
3x^2 + 12x - 18
------------------- = 0
x + 7
2) 9^x + (4)3^+x - 3 = 0

3) 5^x = 4^x+5

4) 2log7(x+6) - log7(x+5) = 2 (the 7's are the log base)

5) log8(e^x+4) = 2 (8is the log base)

Rep and thanks to anyone who will solve these!

2. (1)

$\frac{3x^2 + 12x - 18}{x + 7} = 0$

$\frac{3(x^2 + 4x - 6)}{x + 7} = 0$

To solve this, solve $(x^2 + 4x - 6) = 0$. This gives the only two solutions (use the quadratic formula)

(2)Not sure it isn't written clear enough

(3) Not sure on this one sorry...

(4)

$2\log_{7} (x + 6) - \log_{7}{(x + 5)} = 2$

$\log_{7} [(x + 6)^2] - \log_{7} (x + 5) = 2$

$\log_{7} \left(\frac{(x + 6)^2}{x + 5}\right) = 2$

$\frac{(x + 6)^2}{x + 5} = 7^2 = 49$

$(x + 6)^2 = 49(x + 5)$

Now this is a quadratic equation, solve for x. Remember to check the solutions so that $x + 6 > 0$ and $x + 5 > 0$ because you cannot have logs of negative numbers.

(5)

$\log_{8} (e^x + 4) = 2$

$e^x + 4 = 8^2 = 64$

$e^x = 60$

$x = \log_{e} 60$

3. Thanks for the help!

One small problem, due to me not knowing how to make the problems look neat..

number 5 is

log8(e^(x+4))=2

the e is raised to the x+4..

and on number 2

9^x + (4)3^+x - 3 = 0

it is 9 raised to the x, plus 4 times 3raised to the +x, -3 = 0, if that helps...

thanks a lot again!

4. I'm not too sure on the equations with different numbers raised to power x.

But number (5):

$\log_{8} (e^{x + 4}) = 2$

$e^{x + 4} = 8^2 = 64$

$x + 4 = \log_{e} 64 = \ln 64$