Solve Logarithm Problem: Prove 3/2

  • Thread starter Thread starter chwala
  • Start date Start date
  • Tags Tags
    Logarithm
AI Thread Summary
The discussion revolves around proving the logarithmic equation Log(√27 + Log √8 - Log √125) / (Log 6 - Log 5) = 3/2. Participants emphasize the importance of correctly applying logarithmic properties, such as log(ab) = log(a) + log(b) and log(a^b) = b log(a). A key point made is that the logarithm's base does not affect the outcome of the proof. After simplifying the expression, it is shown that the equation holds true, confirming that the result is indeed 3/2. The conversation highlights the need for clarity in mathematical expressions and the application of logarithmic laws.
chwala
Gold Member
Messages
2,827
Reaction score
415
I am not very good in logs i have tried to attempt below problem in vain..give me an insight.
Prove Log(√27 + Log √8 - Log √125 ) / (Log 6 – Log 5) =3/2
I am not good in logarithms how do you show this…
This is how I did it:,
Log ( ( 3√3 *2√2 ) / 5√5 ) ) / ( 6/5)

((3√(3 *) 2√2 *5)) / ((5√5 *6)) = 6/5 and not 3/2 as indicated by problem am i right?
 
Last edited:
Mathematics news on Phys.org
Is it log10 or the natural logarithm?

Are you sure the problem is correctly written in your post. If you include more parentheses it would be somewhat clearer.

You are expected to use:

log(ab) = log(a) + log(b)
log(a^b) = b log(a)

as well as some factorizing and shortening of fractions.

I checked the numerical value by typing it into Octave, but didn't get 3/2 or 6/5, not with log10 and not with the natural logarithm :-) I won't garantee that I did nothing wrong, though.
 
torquil said:
Is it log10 or the natural logarithm?

Are you sure the problem is correctly written in your post. If you include more parentheses it would be somewhat clearer.

You are expected to use:

log(ab) = log(a) + log(b)
log(a^b) = b log(a)

as well as some factorizing and shortening of fractions.

I checked the numerical value by typing it into Octave, but didn't get 3/2 or 6/5, not with log10 and not with the natural logarithm :-) I won't garantee that I did nothing wrong, though.

This is a logarithms used in every day calculations i.e of base 10 and they are reffered to as common logarithms and the base 10 is often omitted...kindly note that the question is to be solved without the help of tables or other...the solution to the problem is 3/2
kindly solve manually showing your steps i have shown mine...i am conversant with the laws of logarithms.
 
chwala said:
I am not very good in logs i have tried to attempt below problem in vain..give me an insight.
Prove Log(√27 + Log √8 - Log √125 ) / (Log 6 – Log 5) =3/2
I am not good in logarithms how do you show this…
This is how I did it:,
Log ( ( 3√3 *2√2 ) / 5√5 ) ) / log ( 6/5)


Fixed.

You're trying to solve \frac{\log\left(\frac{3\sqrt{3}\cdot 2\sqrt{2}}{5\sqrt{5}}\right)}{\log\left(\frac{6}{5}\right)}



chwala said:
((3√(3 *) 2√2 *5)) / ((5√5 *6)) = 6/5 and not 3/2 as indicated by problem am i right?

\frac{log(a)}{log(b)}\neq \frac{a}{b} so I'm not surprised it doesn't give you the answer you were looking for.

Look at the numerator, simplify the part inside the log, and then convert the roots into fractional powers to make things clearer :wink:

p.s. It doesn't matter what logarithm it is. The answer is independent of this.
 
  • Like
Likes chwala
Log (3^3/2 * 2^3/2 / 5^3/2) = Log (6/5)

Log (3*2/5)^3/2 = Log (6/5)

Log (6/5)^3/2 = Log (6/5)
3/2 Log (6/5) = Log (6/5)
3/2 = log (6/5) / Log (6/5)
=3/2
I love physics forum thank you so much…….maths for life
thanks a lot mentallic.
 
No worries :smile:
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top