Question

Log6^(2x+6)=log36(4x).Решите это уравнение

Answer 1

$log_6(2x+6)=log_{36}(4x) \\ \\ ODZ:x\in (0;+\infty) x^{2} x^{2} \\ \\ log_6(2x+6)=log_{6^{2}}(4x) \\ \\ log_6(2x+6)= \frac{1}{2} *log_6(4x) \\ \\ log_6(2x+6)=log_6((4x)^{ \frac{1}{2}}) \\ \\ 2x+6= \sqrt{4x} \\ \\ 2 \sqrt{x}=2x+6 \\ \\ \sqrt{x} =x+3 \\ \\ x=x^2+6x+9 \\ \\ x^2+5x+9=0 \\ \\ D\ \textless \ 0 \\ \\ x\notin R$

Answer 2

To solve the equation $\log_6^{2x+6} = \log_{36}(4x)$, we'll first use logarithm properties to simplify the equation. Remember that $\log_a^n$ is the logarithm of $n$ to the base $a$.

Step 1: Apply the logarithm property $\log_a^n = \frac{{\log_b^n}}{{\log_b{a}}}$ to both sides of the equation. We can use any base, but let's use the natural logarithm (base $e$) for this demonstration.

$\log_6^{2x+6} = \frac{{\ln(2x+6)}}{{\ln(6)}}$

$\log_{36}(4x) = \frac{{\ln(4x)}}{{\ln(36)}}$

Step 2: Now we have the equation in terms of natural logarithms:

$\frac{{\ln(2x+6)}}{{\ln(6)}} = \frac{{\ln(4x)}}{{\ln(36)}}$

Step 3: Use the property of logarithms $\log_b(a^n) = n\log_b(a)$ to simplify the equation further.

$\frac{{\ln(2x+6)}}{{\ln(6)}} = \frac{{\ln(4) + \ln(x)}}{{\ln(6) + \ln(6)}}$

Step 4: Eliminate the denominators by multiplying both sides by $\ln(6)$.

$\ln(2x+6) = \ln(4) + \ln(x)$

Step 5: Apply the logarithm property $\ln(a) + \ln(b) = \ln(a \cdot b)$.

$\ln(2x+6) = \ln(4x)$

Step 6: Now, remove the natural logarithm by exponentiating both sides with the base $e$.

$e^{\ln(2x+6)} = e^{\ln(4x)}$

Step 7: The exponential and logarithmic functions cancel each other out, leaving us with:

$2x+6 = 4x$

Step 8: Solve for $x$:

$2x = 6$

$x = 3$

Thus, the solution to the equation is $x = 3$.