1. ## log

i have a final exam tomorrow for calculus, and I saw some of the questions
I need help figuring out the answers!

here aree some

1.)condense into a single log:
log5+3log x-2+log y+ (2/3)log z

2.) expand :
ln 4 [cube root of] x / y^3

3.) find the horizontal asymptope of:
f(x)=(4x^2 -7x +6/6x^2 -9)

4.) which quadratic equation has a vertex at (3, -4) and passes through the point (-2, 1)?

i will do ANYTHING for these answers

2. Originally Posted by sofreshsoc1ean
1.)condense into a single log:
log5+3log x-2+log y+ (2/3)log z
1. $\log(a)+\log(b)=\log(a \times b)$

2. $\log(a)-\log(b)=\log(a/ b)$

3. $n \log(a)=\log(a^n)$

4. $n = b^{\log(n)}$ where $b$ is the base of the logarithms in use (here I will assume base 10)

5. $n = \log(b^n)$

So step 1 reduce all the terms in your expression to logs of something:

$\log(5)+3\log (x)-2+\log (y)+ (2/3) \log(z)=$ $
\log(5)+\log (x^3)-\log(10^2)+\log (y)+ \log(z^{2/3})
$

Now collect them together:

$
\log(5)+\log (x^3)-\log(10^2)+\log (y)+ \log(z^{2/3})
$
$
=\log \left( \frac{5 \times x^3 \times y \times z^{2/3} }{10^2} \right)
$

RonL

3. Originally Posted by sofreshsoc1ean
i have a final exam tomorrow for calculus, and I saw some of the questions
I need help figuring out the answers!
[snip]
3.) find the horizontal asymptope of:
f(x)=(4x^2 -7x +6/6x^2 -9)

4.) which quadratic equation has a vertex at (3, -4) and passes through the point (-2, 1)?

i will do ANYTHING for these answers Mr F asks: Really ....?
3.) $f(x) = \frac{\frac{2}{3} (6x^2 - 9) + 12 - 7x}{6x^2 - 9} = \frac{\frac{2}{3} (6x^2 - 9)}{6x^2 - 9} + \frac{12 - 7x}{6x^2 - 9} = \frac{2}{3} + \frac{12 - 7x}{6x^2 - 9}$.

So the horizontal asymptote is y = ........

4.) General turning point form: $y = a(x - h)^2 + k$.

The turning point (vertex) is at (h, k). If the vertex is known, it can be plugged straight in. Then to find a, substitute a known point and solve for the value of a .....

4. Originally Posted by sofreshsoc1ean
3.) find the horizontal asymptope of:
f(x)=(4x^2 -7x +6/6x^2 -9)
we have rational function made up of two polynomials, the highest powers of the numerator and denominator being the same. there are several ways to show that the infinite limits (which is what you need to take to find the horizontal asymptote) is the ratio of the leading coefficients.