Logarithm

[Sultan101]

Is it possible that 5 log5-2 log5 =3log 5

 

[tkhunny]

Why would the difference result in anything else?

 

[pka]

Is it possible that 5 log5-2 log5 =3log 5

\(5\log(5)-2\log(5)=\log\left(5^5\right)-\log\left(5^2\right)=\log\left(\dfrac{5^5}{5^2}\right)=\log\left(5^3\right)=3\log(5)\)

 

[Steven G]

You need to understand what a variable is and think about it at times like this.

I am sure that you would say that 5x-2x = 3x. Well what does that mean? It means that if you replace x with any number then that equation will hold true. For whatever reason x is being replaced with log5. So yes, 5log5 - 2 log5 = 3log5.
The only time this would not be true is if the value of x was not a real number like 12/0.

What if the problem had been 5*7 - 2*7. Wouldn't that be 3*7. You do not have to compute 5*7 and 3*7 to get 3*7 = 21.

Here is a question for you. So please respond back with your answer and steps in solving it. Compute 107*8 - 7*8 in your head. It is extremely simple if you think about it the right way. Go for it!

 

[Sultan101]

Why would the difference result in anything else?

Probably i couldn't make y'all understand what i was about to ask.

 

[tkhunny]

Probably i couldn't make y'all understand what i was about to ask.

Well, feel free to offer all necessary information up front. Don't be shy. The more of your own work you show, the better we can respond.

 

[Sultan101]

Well, feel free to offer all necessary information up front. Don't be shy. The more of your own work you show, the better

 

[Sultan101]

Please help First math is done by me
In the first math when i subtracted the 5log5-2log5 i solved for simple subtraction then with the help logarithmic rules i transferred the coefficient into power .Here i considered log5 as a variable and as we subtract 5a-2a=3a i followed the same rule.
Now my question is is it allowed to consider log as a variable and subtract as variables as there's no rule for subtraction in logarithmic rules
And if there is not then in my 2nd image(red marked) how did they solve it ?

 

[Harry_the_cat]

You're not considering log 5 as a variable. It's not a variable (it doesn't "vary"). You are applying the distributive law.
Is it any different to say 3 sinA + 5 sinA = 8 sin A OR \(\displaystyle 3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}\) ?

 

[HallsofIvy]

Please help First math is done by me
In the first math when i subtracted the 5log5-2log5 i solved for simple subtraction then with the help logarithmic rules i transferred the coefficient into power .Here i considered log5 as a variable and as we subtract 5a-2a=3a i followed the same rule.
Now my question is is it allowed to consider log as a variable and subtract as variables as there's no rule for subtraction in logarithmic rules[/tex]
?? There is log(a)- log(b)= log(a/b).

And if there is not then in my 2nd image(red marked) how did they solve it ?

The part marked in red is \(\displaystyle 3 log(2)- \frac{3}{2}log(2)\). As others have said, that has nothing to do with "properties of logarithms". For any quantities ax- bx= (a- b)x. \(\displaystyle 3x-\frac{3}{2}x= \left(3- \frac{3}{2}\right)x= \frac{3}{2}x\). If "x" happens to be log(2) then \(\displaystyle 3log(2)- \frac{3}{2}log(2)= \left(3- \frac{3}{2}\right)log(2)= \frac{3}{2}log(2)\).