Log Question (1 Viewer)

Arucerious · Jan 31, 2016

InteGrand · Jan 31, 2016

Arucerious said:

$\noindent Using the change of base law,$

$\begin{align*}\log _{a^x}b &=\frac{\log _a b}{\log _a \left(a^x\right)} \\ &= \frac{\log_a b}{x} \quad (\text{as }\log_a \left(a^x\right) = x).\end{align*}$

Nailgun · Jan 31, 2016

$\log _{ { a }^{ x } }{ b } =\frac { \log _{ a }{ b } }{ \log _{ a }{ { a }^{ x } } } =\frac { \log _{ a }{ b } }{ x\log _{ a }{ { a } } } =\frac { \log _{ a }{ b } }{ x }$

edit: ninjaed lel, integrand is a latex cowboy

Drsoccerball · Jan 31, 2016

Nailgun said:
$\log _{ { a }^{ x } }{ b } =\frac { \log _{ a }{ b } }{ \log _{ a }{ { a }^{ x } } } =\frac { \log _{ a }{ b } }{ x\log _{ a }{ { a } } } =\frac { \log _{ a }{ b } }{ x }$

edit: ninjaed lel, integrand is a latex cowboy

$You get use to it quickly and it's like typing normally.$

Arucerious · Jan 31, 2016

Last question, how do I solve this?

If the base were the same, I'd know how to do the question. The part I don't understand is how to get the base to become the same.

Nailgun · Jan 31, 2016

Change of base, with changing base to a

Drsoccerball · Jan 31, 2016

Arucerious said:
Last question, how do I solve this?

If the base were the same, I'd know how to do the question. The part I don't understand is how to get the base to become the same.

$You write it in it's normal form:$

$\frac{\log{(\frac{x+2}{2})}}{\log{\sqrt{a}}} = \frac{\log{(2x)}}{\log{a}}$

$$ Using the fact that: $ \log{a^x} = x \log{a}$

$\frac{2\log{(\frac{x+2}{2})}}{\log{a}}=\frac{\log{(2x)}}{\log{a}}$

$\frac{(x+2)^2}{4}=2x $ Taking note that $ x > 0$

$Solve normally$

$x=2$

Nailgun · Jan 31, 2016

Drsoccerball said:
$You write it in it's normal form:$

$\frac{\log{(\frac{x+2}{2})}}{\log{\sqrt{a}}} = \frac{\log{(2x)}}{\log{a}}$

$$ Using the fact that: $ \log{a^x} = x \log{a}$

$\frac{2\log{(\frac{x+2}{2})}}{\log{a}}=\frac{\log{(2x)}}{\log{a}}$

$\frac{(x+2)^2}{4}=2x $ Taking note that $ x > 0$

$Solve normally$

When you write working in an exam, do you have to note that x>0?

Drsoccerball · Jan 31, 2016

Nailgun said:
When you write working in an exam, do you have to note that x>0?

I would personally just incase one of the answers give you negative. Some teachers if one of the answers is negative and you cross it out but there's no reasoning they will take off a mark.

Arucerious · Feb 1, 2016

Been stuck on this one for awhile, thanks. 10) D

InteGrand · Feb 1, 2016

Arucerious said:
Been stuck on this one for awhile, thanks. 10) D

$\noindent Let $y =\sqrt{x}\log x$, then by the product rule,$

$$\begin{align*} y^\prime &= \frac{1}{2\sqrt{x}}\log x + \sqrt{x}\cdot \frac{1}{x} \\ &= \frac{1}{2\sqrt{x}}\log x + \frac{1}{\sqrt{x}} \\ &= \frac{\frac{1}{2}\log x +1}{\sqrt{x}}.\end{align*}$

Arucerious · Feb 8, 2016

Hi guys, I have another question. Here it is.
Write down the equation of the tangent to y = (log(x))^2 at the point where x = c. Hence find any values of c for which the tangent passes through the origin. I keep arriving at c = 2 but the answer says it is c = 1. Can anyone explain?

InteGrand · Feb 8, 2016

Arucerious said:
Hi guys, I have another question. Here it is.
Write down the equation of the tangent to y = (log(x))^2 at the point where x = c. Hence find any values of c for which the tangent passes through the origin. I keep arriving at c = 2 but the answer says it is c = 1. Can anyone explain?

$\noindent Let $y=f(x)=\left( \ln x\right)^2$ ($x>0$), then $f^\prime (c) = 2\ln c \cdot \frac{1}{c}$. The point on the curve at $x = c$ is $\left(c, \left(\ln c\right)^2 \right)$. So using point-slope form, the equation of the tangent to the curve here is $y -\left(\ln c\right)^2 = \frac{2\ln c}{c} \left(x -c\right)$. Rearranging, we see that the $y$-intercept is $-2\ln c +\left(\ln c \right)^2$.$

$\noindent For the tangent to pass through the origin, we want this $y$-intercept to be 0. So we solve $-2\ln c + \left(\ln c\right)^2 = 0$. Factorising, $\ln c \left(-2 + \ln c\right)=0$. So either $\ln c = 0\Longleftrightarrow c =1$ or $\ln c -2 =0 \Longleftrightarrow c =\mathrm{e}^{2}$. So these are the two values of $c$ that give a tangent that goes through the origin.$

Arucerious · Feb 8, 2016

Thanks so much, I was thinking what went wrong for ages but seems I started messing up at the point where I reached -(ln(c))^2 = -2ln(c)

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