1. ## Logarithem equation

Find the value of X such that:

log2x + log232 = 6logx2

2. $\displaystyle \log _2 x + \log _2 32 = 6\log _x 2 \hfill \\$

$\displaystyle \Rightarrow \log _2 \left( {x \times 32} \right) = 6\left( {\frac{{\log 2}} {{\log x}}} \right) \hfill$

$\displaystyle \Rightarrow \log _2 \left( {32x} \right) = \frac{{6\log 2}} {{\log x}} \hfill \\$

$\displaystyle \Rightarrow \log _2 \left( {32x} \right) = \frac{{\log \left( 2 \right)^6 }} {{\log x}} \hfill \\$

$\displaystyle \Rightarrow \frac{{\log \left( {32x} \right)}} {{\log 2}} = \frac{{\log 64}} {{\log x}} \hfill \\$

$\displaystyle \Rightarrow x = 2$

3. thx, but I did not understand how u got x=2 from

4. Hello,
Originally Posted by Musab
thx, but I did not understand how u got x=2 from
By observation.

Or another method :
$\displaystyle \log_2(x)+\log_2(32)=6 \log_x(2)$

32=2^5 ---> $\displaystyle \log_2(32)=5$

$\displaystyle \frac{\ln(x)}{\ln(2)}+5=6 \frac{\ln(2)}{\ln(x)}$

Multiply the two sides by $\displaystyle \ln(2) \cdot \ln(x)$ and solve the quadratic

5. Hello, Musab!

Find the value of x such that: .$\displaystyle \log_2(x) + \log_2(32) \:=\:6\log_x(2)$
Recall the theorem: .$\displaystyle \log_a(b)) \:=\:\frac{1}{\log_b(a)}$

We have: .$\displaystyle \log_2(x) + 5 \;=\;\frac{6}{\log_2(x)}$

Then: .$\displaystyle \left[\log_2(x)\right]^2 + 5\left[\log_2(x)\right] \:=\:6 \quad\Rightarrow\quad [\log_2(x)]^2 + 5\left[\log_2(x)\right] - 6 \:=\:0$

Factor: .$\displaystyle \left[\log_2(x)+6\right]\,\left[\log_2(x) - 1\right] \:=\:0$

Then: .$\displaystyle \log_2(x) +6 \:=\:\quad\Rightarrow\quad \log_2(x) \:=\:-6 \quad\Rightarrow\quad x \:=\:2^{-6} \quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{64}}$

And: .$\displaystyle \log_2(x) -1 \:=\: \quad\Rightarrow\quad \log_2(x) \:=\:1\quad\Rightarrow\quad x \:=\:2^1 \quad\Rightarrow\quad \boxed{x\:=\:2}$

Edit: corrected an awful blunder . . .
.