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

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- May 21st 2006, 11:27 PMkingkaisai2Difficult simplification for logarithm
How do you simplify log(2square root of 10)-1/3 log0.8-log10/3

- May 22nd 2006, 06:13 AMticbolQuote:

Originally Posted by**kingkaisai2**

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)

= Log(3) -(1/6) ----------------------------answer.

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, 05:36 PMSoroban
Hello, kingkaisai2!

I assume I'm reading it correctly . . .

Quote:

Simplify:

Then:

And:

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

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

The denominator is:

. . .

The fraction becomes:

Rationalize: