# Difficult simplification for logarithm

• May 21st 2006, 10:27 PM
kingkaisai2
Difficult simplification for logarithm
How do you simplify log(2square root of 10)-1/3 log0.8-log10/3
• May 22nd 2006, 05:13 AM
ticbol
Quote:

Originally Posted by kingkaisai2
How do you simplify log(2square root of 10)-1/3 log0.8-log10/3

Log[2sqrt(10)] -(1/3)Log(0.8) -Log(10/3) -------(i)

Is that what you mean?
If yes, do you want to combine the 3 logs into one log only?
If yes again, then,
= Log[2sqrt(10)] -Log[cubrt(0.8)] - Log(10/3)
= Log{[2sqrt(10)] / [cubrt(0.8)] / (10/3)} ----------(ii)

Umm, it's getting too complicated. Let us simplify the 3 logs separately, for simplicity.
I assume, by "simplify", you mean no decimal points.

Log[2sqrt(10)]
= Log[2*(10)^(1/2)]
= Log(2) +Log(10^1/2)
= Log(2) +(1/2)Log(10)
= Log(2) +(1/2)(1)
= Log(2) +(1/2) ------**

(1/3)Log(0.8)
= (1/3)Log[8/10]
= (1/3)[Log(8) -Log(10)]
= (1/3)Log(8) -(1/3)Log(10)
= (1/3)Log(2^3) -(1/3)(1)
= Log[(2^3)^(1/3)] -(1/3)
= Log[2] -(1/3) ----------**

Log(10/3)
= Log(10) -Log(3)
= 1 -Log(3) ----------------**

Therefore, (i) becomes
= [Log(2) +(1/2)] -[Log(2) -(1/3)] -[1 -Log(3)]
= 1/2 +1/3 -1 +Log(3)
= (3 +2 -6)/6 +Log(3)
= -1/6 +Log(3)

Check,
Using a calculator,

Log[2sqrt(10)] -(1/3)Log(0.8) -Log(10/3)
= 0.80103 -(-0.03230) -0.52288
= 0.31045

Log(3) -1/6
= 0.47712 -0.16667
= 0.31045

They are the same, so, OK.
• May 28th 2006, 04:36 PM
Soroban
Hello, kingkaisai2!

I assume I'm reading it correctly . . .

Quote:

Simplify: $\displaystyle \log(2\sqrt{10}) - \frac{1}{3}\log(0.8) -\log\left(\frac{10}{3}\right)$
We have: $\displaystyle \log(2\sqrt{10}) - \log(0.8^{\frac{1}{3}}) - [\log(10) - \log(3)]$

Then: $\displaystyle \log(2\sqrt{10}) - \log(\sqrt[3]{0.8}) - \log(10) + \log(3)$

And: $\displaystyle \log\left[\frac{2\sqrt{10}\cdot3}{\sqrt[3]{0.8}\cdot10}\right] \;= \;\log\left[\frac{3}{\sqrt{10}\cdot\sqrt[3]{0.8}}\right]$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you really want to impress/surprise/terrify your teacher, keep going.

The denominator is: $\displaystyle \sqrt{10}\cdot\sqrt[3]{8\cdot10^{-1}} \;= \;10^{\frac{1}{2}}\cdot\sqrt[3]{8}\cdot10^{-\frac{1}{3}}$

. . . $\displaystyle = \;2\left(10^{\frac{1}{2}}\cdot10^{-\frac{1}{3}}\right) \;= \;2\cdot10^{\frac{1}{6}}$

The fraction becomes: $\displaystyle \frac{3}{2\cdot10^{\frac{1}{6}}}$

Rationalize: $\displaystyle \frac{3}{2\cdot10^{\frac{1}{6}}} \cdot \frac{10^{\frac{5}{6}}}{10^{\frac{5}{6}}} \;=\;\frac{3\cdot10^{5/6}}{20}$