Question

Log x(4)=2 log x-3 (27)=3

Answer 1

Ответ:

Объяснение:

x^2=4,  x=2,  x=-2,  но x>0,  отв. х=2

x-3>0, x>3,    (x-3)^3=27, извлекаем корень кубический из обеих

частей,  x-3=3,  x=6   ( ^-знак степени)

Answer 2

Ответ:

Объяснение:

$\log_x4=2\\\dfrac{\log_24}{\log_2x}=2\\\log_24=2\cdot\log_2x\\2=2\cdot\log_2x\\\log_2x=1\\2^{\log_2x}=2^1\\x=2\\\\\log_{x-3}27=3\\\dfrac{\log_327}{\log_3(x-3)}=3\\\log_327=3\cdot\log_3(x-3)\\\log_33^3=3\cdot\log_3(x-3)\\3=3\cdot\log_3(x-3)\\\log_3(x-3)=1\\3^{\log_3(x-3)}=3^1\\x-3=3\\x=6$

Answer 3

It seems like you have a couple of logarithmic equations involving the base "x." Let's solve them step by step.

1. Equation 1: $\log_x(4) = 2 \log_x - 3 (27) = 3$

2. Solving Equation 1: Let's start by simplifying the logarithmic expressions:

   $\log_x(4) = 2 \log_x - 3 (27)$

   First, simplify the right side by applying the power rule of logarithms and evaluating $3 \cdot 27$:

   $\log_x(4) = 2 \log_x(3^3)$

   Now, use the logarithmic identity $\log_a(b^c) = c \log_a(b)$:

   $\log_x(4) = 6 \log_x(3)$

3. Equation 2: $\log_x(4) = 6 \log_x(3)$

4. Solving Equation 2: We have the same logarithmic expression on both sides, which means the bases must be equal. Therefore, we can set up an equation:

   $4 = 3^6$

   Simplify the right side:

   $4 = 729$

   However, this is not true, which means there might be an issue in the initial equations or calculations.

   Double-check the original equations and calculations to ensure their accuracy. If you're able to provide the correct equations, I'd be happy to help you solve them.