How can I evaluate this?

[Chikis]

[MATH]\frac{3.87^3+20}{3.87^3-20}[/MATH]

I want solve the above using logarithm. I don't know how to solve it using logarithm because I can't deal with this: [MATH]\frac{10^{(0.5877)3}+10^{1.3010}}{10^{(0.5877)3}-10^{1.3010}}[/MATH]

If this is given: [MATH]\frac{3.87^3\times20}{3.87^3\times20}[/MATH] It would have been easiar for me to solve if I add and subtract all the powers that has base 10. Please I need help.

[studiot]

What is 20 the log of?

 

and what is 3.87³ the log of?

 

And what is the log of A added to the log of B the log of?

[Chikis]

But[MATH]\frac{3.87^3+20}{3.87^3-20}[/MATH]

=[MATH]\frac{10^{(0.5877)3}+10^{1.3010}}{10^{(0.5877)3}-10^{1.3010}}[/MATH]

What else can I do?

Edited February 19, 2015 by Chikis

[Chikis]

Please come to my aid. I have shown my work. I can't make any headway without help.

[Nicholas Kang]

You should have solved the problem with ease by simply using normal mathematics, bypassing logarithm. Does the question state that logarithm must be used? From post #3, simply add 0.5877(3) and 1.3010 for the upper part and 0.5877(3) minus 1.3010 for the bottom part.

 

You get 10 to the power of [0.5877(3)+1.3010] divides by 10 to the power of [0.5877(3)-1.3010]

 

Then, further simplify the equation and you get 10 to the power of {[0.5877(3)+1.3010]-[0.5877(3)-1.3010]}

 

I can only help with high school maths. Sorry.

The answer is approximately 400. Try and see.

[Chikis]

The answer is incorrect![MATH]\frac{3.87^3+20}{3.87^3-20}[/MATH] = 2.05

I want to solve the problem using logathim and still have the same as when calculating it ordinarly.

Edited February 20, 2015 by Chikis

[ajb]

The answer is approximately 400. Try and see.

How did you get 400? A simple approximation of 3.87 as 4 shows that the value is close to 2 (as Chikis states) and not close to 400.

[studiot]

I think I originally misunderstood you.

 

Why will taking the log(77.960603) and subtracting log(37.960603) and then taking the antilog not give your answer?

[ajb]

Why will taking the log(77.960603) and subtracting log(37.960603) and then taking the antilog not give your answer?

It will of course. I was wondering if the OP was looking for some clever simplification here. I cannot see anything past your suggestion. The log is not not linear so nothing nice will happen with the sum.