[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.

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?

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, 201510 yr by Chikis

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

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.

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, 201510 yr by Chikis

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.

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?

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.