Question

log2 (x)+ logx (4)=3

Answer 1

i can't understand with this

Answer 2

Trial and Error seems to be the best solution to this. If you're familiar with the property
\(\log_b(b) = 1\) that should help you.

Answer 3

\[\log _x 4 = 2 \log _x 2\]Also notice that,\[\log_x 2 = \dfrac{1}{\log_2 x}\]

Answer 4

assume log log 2 x=t

Answer 5

t+2/t=3
now solve for t

Answer 6

then check the solution in the equation

Answer 7

t^2-3t+2=0

Answer 8

t=2,1

Answer 9

log 2 x=2=>x=4
log 2 x=1 => x=2

Answer 10

\[ \large \log_2x+\log_x4=3\\
\large \frac{\log x}{\log2}+\frac{\log 4}{\log x}=3\\
\large \frac{(\log x)^2}{\log 2}+\log 4= 3 \log x\\
\large \frac{(\log x)^2+(\log 2)(\log 4)}{\log 2}= 3 \log x\\
\large (\log x)^2-(3\log 2)(\log x)+(\log 2)(\log 4)=0\\
\large (\log x)^2-(3\log 2)(\log x)+(\log2)(\log2^2)=0\\
\large (\log x)^2-(3\log 2)(\log x)+2(\log 2)^2=0\], since \(\log 2^2= 2 \log 2\)
You can think of this a quadratic if you replace \(y=\log x\)
\[\large y^2-3\log 2(y)+2(\log 2)^2=0 \]
Quadratic formula:
\[\large y=\frac{-(-3 \log 2)\pm \sqrt{(-3 \log 2)^2-4(1)(2(\log2)^2)}}{2(1)} \\
\large y=\frac{3 \log 2 \pm \sqrt{9(\log 2)^2 -8(\log 2)^2}}{2}\\
\large y=\frac{3 \log 2 \pm \sqrt{(\log 2)^2}}{2}\\
\large y = \frac{3 \log 2 \pm \log 2}{2}\\
\large y = \frac{4 \log 2}{2}=2 \log 2, \text{ or } \frac{2 \log 2}{2} = \log 2\]
Now recall \(y = \log x\)
So:
\(\large \log x = 2 \log 2 = \log 2^2 = \log 4\) and \(\large \log x = \log 2\)
Thus
\(\boxed{\large x=2, x=4}\)

Answer 11

cody's solution is where I was getting at... but well...