Question

log9 (3log2 (1 + log3 (1 + 2log2x))) = \\frac{1}{2}\\). Find x.
4
\\frac{1}{2}\\)
1
2

Answer 1

Given:

Log9 (3log2 (1 + log3 (1 + 2log2x))) = frac{1}{2}).  

To find:

Find x.

Solution:

From given we have,

Log9 (3log2 (1 + log3 (1 + 2log2x))) = frac{1}{2}).

log _9(3log _2(1+log _3(1+2\log _2(x)))) = 1/2

3log _2(1+log _3(1+2log _2(x))) = \sqrt{9}

3log _2(1+log _3(1+2log _2(x))) = \sqrt{3^2}

3log _2(1+log _3(1+2log _2(x))) = 3

log _2(1+log _3(1+2log _2(x))) = 1

1+log _3(1+2log _2(x)) = 2^1

log _3(1+2log _2(x))-1 = 2-1

log _3(1+2log _2(x)) = 1

1+2log _2(x) = 3^1

2log _2(x)-1 = 3-1

2log _2(x) = 2

log _2(x) = 1

x = 2^1

x = 2

Therefore, the value of x is 2

$\log _9\left(3\log _2\left(1+\log _3\left(1+2\log _2\left(x\right)\right)\right)\right)=\frac{1}{2}\\3\log _2\left(1+\log _3\left(1+2\log _2\left(x\right)\right)\right)=\sqrt{9}\\3\log _2\left(1+\log _3\left(1+2\log _2\left(x\right)\right)\right)=\sqrt{3^2}\\3\log _2\left(1+\log _3\left(1+2\log _2\left(x\right)\right)\right)=3$

$\log _2\left(1+\log _3\left(1+2\log _2\left(x\right)\right)\right)=1\\1+\log _3\left(1+2\log _2\left(x\right)\right)=2^1\\\log _3\left(1+2\log _2\left(x\right)\right)-1=2-1\\\log _3\left(1+2\log _2\left(x\right)\right)=1$

$1+2\log _2\left(x\right)=3^1\\2\log _2\left(x\right)-1=3-1\\2\log _2\left(x\right)=2\log _2\left(x\right)=1\\x=2^1\\x=2$