It looks like you are changing everything to log base 10. Your workings for changing base are correct but you only need to change one of them, either put 3log_{4}(x) into base x or log_{x}(16) into base 4
3 log base 4 x + log base x 16 = 11/2
First I know I have to use change of base rule
3 log base 4 x will be log x^3/log 4 =3 log x/2 log 2
Log base x 16 will be equal to log 16/log x = 2/log x
But this method is'nt working. How do I manipulate this log equation?
Critism is welcome.
I wl prefer changing to base 4.@ Soroban nt rejecting your solution bt my memo changed the equation to base 4
And it will be 3[ log base 4 (x^2)] + log base (4) 16/log 4 (x) =11/2
Now iam stuck with how the first expression became 3[log base 4 (x^2)]
Infact the (x^2) part
@Bonganitedd What you have written is totally goobly-gook.
If you cannot express yourself in reasonable standard English do not expect help at this site.
We are glad to help persons who are serious and have a basic grasp of basics.
Sadly, that does not seemly apply to you.
Ok I am sory abt dat.
What i was trying to say is, I have a marking memoranda for the above question, but I am lost@ step 2. I will rewrite it.
3 log base(4) x + log base (x) 16 = 11/2(original log equation)
According to my marking memoranda
3 log base (4) x + log base (4) 16/log base (4) x =11/2. (Change the base law)
3[log base (4) x^2] + 2 = 11/2 log base (4) ( this is where I am lost)
Does the equation make sense now?
3 log base (4) x + 2/log base (4) x =11/2 (multiply out by log base (4) x)
Then am having 3 (log base (4) x )(log base(4) x) +2 =log 11/2 log base(4) x
Then 3(log base (4) x)^2 + 2 = 11/2 log base (4) x (multiply by 2)
6(log base (4) x)^2 + 4 =11 log base (4) x
6(log base (4) x)^2 - 11 log base (4) x + 4 = 0
(3 log base (4) x - 4)(2 log base (4) x -1) = 0
Log base (4) x = 4/3 or log base (4) x = 1/2
X = 4^4/3 or x = 2
I learned in hard way.
Much thanks to Shakarri, Plato, Soroban and ibdutt