# Thread: Logarithmic Question.

1. ## Logarithmic Question.

. Mr. Logarithmic is showing his class a solution to a problem. Is Mr. Logarithmic correct? Explain and correct if he is incorrect. Express 2log25 - log2(4/5) + 1/2log216 as a simple logarithm = 2log25 - log2(4/5) + 1/2log216 = log152-log2(4/5) + log2(16/2) = log225 - log2(4/5)+ log2(8) = log2 (25 x 4/5 x 8)

So, from my understanding, the solution is incorrect the correct solution according to my calculations is:'

2log25 - log2(4/5) + 1/2log216
log225 - log2(4/5) + log24
Therefore, we get:

log2 (25/(4/5) X 4)

I'm really confused here. Is my working/solution correct for the simplification of the logarithm? Any helpful tips/suggestions would be greatly appreciated.

Thanks!

2. Originally Posted by spoc21
. Mr. Logarithmic is showing his class a solution to a problem. Is Mr. Logarithmic correct? Explain and correct if he is incorrect. Express 2log25 - log2(4/5) + 1/2log216 as a simple logarithm
l
your notation leaves a bit to be desired ... is this the expression?

$2\log_2{5} - \log_2\left(\frac{4}{5}\right) + \frac{1}{2}\log_2{16}
$

3. Originally Posted by skeeter
your notation leaves a bit to be desired ... is this the expression?

$2\log_2{5} - \log_2\left(\frac{4}{5}\right) + \frac{1}{2}\log_2{16}
$
yes, thats exactly how it is.

4. ## Didn't see the minus sign ... corrected

Hi !
Let's see how it goes.

$2\log_2{(5)} - \log_2{\left(\frac{4}{5}\right)} + \frac{1}{2}\log_2{(16)}$

All the logarithms have the same base, so we won't have to change this. First, simplify any coefficients in front of the logarithms, using the power law :

$\log_2{(5^2)} - \log_2\left(\frac{4}{5}\right) + \log_2{\left (16^{\frac{1}{2}} \right )}$

Now simplify the expressions :

$\log_2{(25)} - \log_2\left(\frac{4}{5}\right) + \log_2{(4)}$

Now it gets tricky : you need brackets otherwise it's not going to work. So put the last two terms into brackets :

$\log_2{(25)} - \left ( \log_2\left(\frac{4}{5}\right) - \log_2{(4)} \right )$

Now use the logarithm division/substraction law on the two last terms :

$\log_2{(25)} - \log_2{\left (\frac{\frac{4}{5}}{4} \right )}$

$\log_2{(25)} - \log_2{\left (\frac{4}{20} \right )}$

Now divide using the logarithm division/substraction law :

$\log_2{\left ( \frac{25}{\frac{4}{20}} \right )}$

$\log_2{(125)}$

$3 \log_2{(5)}$

This is the simplest form of this expression.
____________________________________

Mr. Logarithmic is wrong, and your reasoning is correct although a bit hard to follow without proper mathematical fonts. You should have simplified further to get 125 which will have concluded the exercise.

5. Hello, spoc21!

Mr. Logarithmic is showing his class a solution to a problem.
Is Mr. Logarithmic correct? Explain and correct if he is incorrect.
. . He made two errors.

Express $2\log_2(5) - \log_2(\tfrac{4}{5}) + \tfrac{1}{2}\log_2(16)$ as a simple logarithm.

$2\log_2(5) - \log_2\tfrac{4}{5}) + \tfrac{1}{2}\log_2(16)$

. . $=\;\log_2\left(5^2\right) -\log_2(\tfrac{4}{5}) + \log_2\left({\color{red}\tfrac{16}{2}}\right)$

. . $=\; \log_2(25) - \log_2\left(\tfrac{4}{5}\right)+ \log_2(8)$

. . $=\; \log_2\left(25 \,{\color{red}\times}\, \tfrac{4}{5} \times 8\right)$

$2\log_2(5) - \log_2(\tfrac{4}{5}) + \tfrac{1}{2}\log_2(16)$

. . $=\;\log_2(5^2) - \bigg[\log_2(4)-\log_2(5)\bigg] + \log_2(16^{\frac{1}{2}})$

. . $=\;\log_2(25) - \log_2(4) + \log_2(5) + \log_2(4)$

. . $=\;\log_2(25) + \log_2(5)$

. . $=\;\log_2(25 \times 5)$

. . $=\;\log_2(125)$