another log question (1 Viewer)

Green Yoda · Jul 4, 2015

sorry about the mass amounts of log questions..i have literally no one to consult with in the holidays..

Question 3H?

rand_althor · Jul 4, 2015

photastic · Jul 4, 2015

5(1/2)^x=20
(1/2)^x=4
log sides
ln[(1/2)^x]=ln4
xln(1/2)=ln4
x=ln4/ln(1/2)

or you could equate the indices.

Drongoski · Jul 4, 2015

Green Yoda · Jul 4, 2015

photastic said:
5(1/2)^x=20
(1/2)^x=4
log sides
ln[(1/2)^x]=ln4
xln(1/2)=ln4
x=ln4/ln(1/2)

or you could equate the indices.

can you elaborate on that? full working out?

Green Yoda · Jul 4, 2015

Drongoski said:
$5(\frac {1}{2})^x = 20 \implies (\frac {1}{2})^x = 2^{-x} = \frac {20}{5} = 4 = 2^2 \implies -x = 2 \implies x = -2$

hmm first of all mind my stupidity. I have no idea what you did there...you did that without logs?

rand_althor · Jul 4, 2015

Rathin said:
can you elaborate on that? full working out?

photastic's method:
$\begin{align*}5(\frac{1}{2})^x&=20 \\(\frac{1}{2})^x&=4 \\\textup{Taking the logarithms of both sides} \\\log{\left [(\frac{1}{2})^x \right ]}&=\log{\left [2^2 \right ]} \\\textup{Using the rule }\log_a{\left [m^p \right ]}&=p\log_a{m} \\x\log{\left [\frac{1}{2} \right ]}&=2\log{2} \\x&=\frac{2\log2}{\log{\left [2^{-1} \right ]}} \\&=\frac{2\log2}{-\log2} \\&=\frac{2}{-1} \\&=-2\end{align*}$

Green Yoda · Jul 4, 2015

rand_althor said:
photastic's method:
$\begin{align*}5(\frac{1}{2})^x&=20 \\(\frac{1}{2})^x&=4 \\\textup{Taking the logarithms of both sides} \\\log{\left [(\frac{1}{2})^x \right ]}&=\log{\left [2^2 \right ]} \\\textup{Using the rule }log_a{\left [m^p \right ]}&=p\log_a{m} \\x\log{\left [\frac{1}{2} \right ]}&=2\log{2} \\x&=\frac{2\log2}{\log{\left [2^{-1} \right ]}} \\&=\frac{2\log2}{-\log2} \\&=\frac{2}{-1} \\&=-2\end{align*}$

omg thank you very much!! also is it still possible to do it with taking logs to both sides from the beginning instead of dividing 5 from both sides?

rand_althor · Jul 4, 2015

Rathin said:
omg thank you very much!! also is it still possible to do it with taking logs to both sides from the beginning instead of dividing 5 from both sides?

$\begin{align*}5(\frac{1}{2})^x&=20 \\\log{\left [5(\frac{1}{2})^x \right ]}&= \log{20} \\\textup{Using the rule }\log{a}+\log{b}&= \log{ab} \\\log{5}+\log{\left [(\frac{1}{2})^x \right ]}&=\log{5}+\log{4} \\\log{\left [(\frac{1}{2})^x \right ]}&=\log{2^2} \\\end{align*}$

And it's the same from there.

Green Yoda · Jul 4, 2015

rand_althor said:
$\begin{align*}5(\frac{1}{2})^x&=20 \\\log{\left [5(\frac{1}{2})^x \right ]}&= \log{20} \\\textup{Using the rule }\log{a}+\log{b}&= \log{ab} \\\log{5}+\log{\left [(\frac{1}{2})^x \right ]}&=\log{5}+\log{4} \\\log{\left [(\frac{1}{2})^x \right ]}&=\log{2^2} \\\end{align*}$

And it's the same from there.

I did a similar thing but instead of keeping x with 1/2 i put it outside of the bracket.. which kinda stuffed me up and also why does this still work with a base of 10 as per log 5? because the question asks us to make the base 2

rand_althor · Jul 4, 2015

Yep that's because $\log{\left [x^2 \right ]}$ is not the same as $\left [\log{x} \right ]^2$ .

As far as I am aware, the base doesn't matter with questions like these. I think their intention with asking you to use a base of 2 was to use the rule $\log_a{\left [a \right ]}=1$

Green Yoda · Jul 4, 2015

rand_althor said:
Yep that's because $\log{\left [x^2 \right ]}$ is not the same as $\left [\log{x} \right ]^2$ .

As far as I am aware, the base doesn't matter with questions like these. I think their intention with asking you to use a base of 2 was to use the identity $\log_a{\left [a \right ]}=1$

other questions were pretty easy but they put this in the last question...so this is like one of a kind in the book. How would when to use the base as the question asks and when not to?

rand_althor · Jul 4, 2015

As you can see from the solutions in this thread, the base wasn't important in achieving the final answer. I guess if you see that the terms in the equation you're trying to solve are powers of a certain number then you could make that number your base. However, I don't think it will matter too much.

Green Yoda · Jul 4, 2015

rand_althor said:
As you can see from the solutions in this thread, the base wasn't important in achieving the final answer. I guess if you see that the terms in the equation you're trying to solve are powers of a certain number then you could make that number your base. However, I don't think it will matter too much.

So any question from question 3 can be done without worrying about the base?

rand_althor · Jul 4, 2015

Yep, I think so.

Green Yoda · Jul 4, 2015

rand_althor said:
Yep, I think so.

Just quickly can you please do this method on any other question in q3 so I can have another example to look at. Also thank you very much for taking your time to help me

anomalousdecay · Jul 4, 2015

Rathin said:
So any question from question 3 can be done without worrying about the base?

Yes. The base does not matter because you are manipulating both sides of the equation. Just make sure that you use the same base of log for both sides and you will be fine.

rand_althor · Jul 4, 2015

Rathin said:
Just quickly can you please do this method on any other question in q3 so I can have another example to look at. Also thank you very much for taking your time to help me

No worries, I'm glad to be of assistance. Here is 3b:

$\begin{align*}8^x&=32\\\log{\left [8^x \right ]}&=\log{32}\\x\log{\left [2^3 \right ]}&=\log{\left [2^5 \right ]} \\3x\log{2}&=5\log{2} \\x&=\frac{5}{3}\\\end{align*}$

Green Yoda · Jul 4, 2015

rand_althor said:
No worries, I'm glad to be of assistance. Here is 3b:
$\begin{align*}8^x&=32\\ \log{\left [8^x \right ]}&=\log{32}\\ x\log{\left [2^3 \right ]}&=\log{\left [2^5 \right ]} \\3x\log{2}&=5\log{2} \\x&=\frac{5}{3}\\ \end{align*}$

You had to put the base value here as 2 right?

InteGrand · Jul 4, 2015

Rathin said:
You had to put the base value here as 2 right?

No, all the log laws he used there work for any log base.

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