\(5\log(5)-2\log(5)=\log\left(5^5\right)-\log\left(5^2\right)=\log\left(\dfrac{5^5}{5^2}\right)=\log\left(5^3\right)=3\log(5)\)Is it possible that 5 log5-2 log5 =3log 5
Probably i couldn't make y'all understand what i was about to ask.Why would the difference result in anything else?
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.Probably i couldn't make y'all understand what i was about to ask.
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 \(\displaystyle 3log(2)- \frac{3}{2}log(2)= \left(3- \frac{3}{2}\right)log(2)= \frac{3}{2}log(2)\).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 ?