Question

Log4 x^2 + logx^6 64 = 2

Answer 1

x≠1; x>0;

$\displaystyle log_{2^2} x^2 + log_{x^6}2^6 = 2;\\log_2x + log_x2 = 2;\\log_2x + \frac{1}{log_2x} = 2;\ *log_2x;\ log_2x\neq 0; x\neq 1;\\log^2_2x + 1 - 2log_2x=0;\ t=log_2x;$

t²-2t+1=0;

t=4-4=0;

t=(2)/2=1;

log₂x=1;

x=2;

Answer 2

To solve the equation $\log_4(x^2) + \log_x^6(64) = 2$, we can use the properties of logarithms. First, let's simplify the expression:

1. Use the property of logarithms that says $\log_a(b^c) = c \cdot \log_a(b)$:

   $2 \cdot \log_4(x) + 6 \cdot \log_x(64) = 2$

2. Next, simplify $\log_x(64)$. Since $64 = x^6$, we can write it as $x^6$:

   $2 \cdot \log_4(x) + 6 \cdot \log_x(x^6) = 2$

3. Now, use the property $\log_a(a) = 1$ to simplify $\log_x(x^6)$:

   $2 \cdot \log_4(x) + 6 \cdot 1 = 2$

4. Simplify further:

   $2 \cdot \log_4(x) + 6 = 2$

5. Subtract 6 from both sides to isolate the logarithmic term:

   $2 \cdot \log_4(x) = 2 - 6$

6. Combine the constants on the right side:

   $2 \cdot \log_4(x) = -4$

7. Divide both sides by 2:

   $\log_4(x) = -2$

Now, we need to solve for $x$. To do that, we can rewrite the equation in exponential form:

$4^{-2} = x$

$\frac{1}{4^2} = x$

$\frac{1}{16} = x$

So, the solution to the equation is $x = \frac{1}{16}$.