Question

Log1/3(4x+1)=-3 please​

Answer 1

$log_{\frac{1}{3}}(4x+1)=-3\\ODZ:x>-\frac{1}{4}\\log_{\frac{1}{3}}(4x+1)=log_{\frac{1}{3}}27\\4x-1=27\\4x=28\\x=7$Ответ:

Пошаговое объяснение:

Answer 2

To solve the equation $\log_{\frac{1}{3}}(4x+1) = -3$, we can follow these steps:

Step 1: Rewrite the equation in exponential form. $\log_{\frac{1}{3}}(4x+1) = -3$ can be rewritten as: $\left(\frac{1}{3}\right)^{-3} = 4x+1$.

Step 2: Simplify the left side. $\left(\frac{1}{3}\right)^{-3} = 3^3 = 27$.

Step 3: Solve for $x$. $27 = 4x + 1$.

Subtract 1 from both sides: $26 = 4x$.

Divide both sides by 4: $x = \frac{26}{4}$.

Simplify: $x = \frac{13}{2}$ or $x = 6.5$.

So, the solution to the equation $\log_{\frac{1}{3}}(4x+1) = -3$ is $x = 6.5$.