Math Help - logarithm manipulation

1. logarithm manipulation

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.

2. Re: logarithm manipulation

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 3log4(x) into base x or logx(16) into base 4

3. Re: logarithm manipulation

Can you help with that step?

4. Re: logarithm manipulation

The general rule to convert from base a to base b is $log_a(y)=\frac{log_b(y)}{log_b(a)}$
So $log_x(16)=\frac{log_4(16)}{log_4(x)}$

5. Re: logarithm manipulation

Hello, Bonganitedd!

$3\log_4\!x + \log_x\!16 \:=\: \tfrac{11}{2}$

I would change everything to base-2.

. . $3\frac{\log_2\!x}{\log_2\!4} + \frac{\log_2\!16}{\log_2\!x} \:=\:\frac{11}{2} \quad\Rightarrow\quad \frac{3\log_2\!x}{2} + \frac{4}{\log_2\!x} \:=\:\frac{11}{2}$

Multiply by $2\log_2\!x\!:$

. . $3(\log_2\!x)^2 + 8 \:=\:11\log_2\!x \quad\Rightarrow\quad 3(\log_2\!x)^2 - 11\log_2\!x + 8 \:=\:0$

Factor: . $(\log_2\!x - 1)(3\log_2\!x - 8) \:=\:0$

Therefore: . $\begin{Bmatrix}\log_2\!x -1 \:=\:0 & \Rightarrow & \log_2\!x \:=\:1 & \Rightarrow & x \:=\:2 \\ 3\log_2\!x -8 \:=\:0 & \Rightarrow & \log_2\!x \:=\:\frac{8}{3} & \Rightarrow & x \:=\:2^{\frac{8}{3}} \end{Bmatrix}$

6. Re: logarithm manipulation

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

7. Re: logarithm manipulation

Originally Posted by Bonganitedd
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.

8. Re: logarithm manipulation

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?

9. Re: logarithm manipulation

I can simplify log base (4) 16/log base (4) x to become 2/log base (4) x
Now I will have 3 log base (4) x + 2/log base (4) x = 11/2
Then I multiply out by log base (4) x

Is this correct?

11. Re: logarithm manipulation

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