Question

Log6 (x+4)+log6 (x-1)=1

Answer 1

$log_{6}(x+4)+log_{6} (x-1)=1 x+4+x-1=1 2x+3=1 2x=1-3 2x=-2 x=-1$

Answer 2

To solve the equation $\log_6(x+4) + \log_6(x-1) = 1$, you can use the properties of logarithms. First, you can combine the two logarithms into one using the properties of logarithms:

$\log_6((x+4)(x-1)) = 1$

Now, you can rewrite this equation in exponential form to solve for (x:

$6^1 = (x+4)(x-1)$

Now, simplify the equation:

$6 = (x+4)(x-1)$

Expand the right side of the equation:

$6 = x^2 + 3x - 4$

Now, bring all terms to one side of the equation to set it equal to zero:

$x^2 + 3x - 4 - 6 = 0$

Simplify further:

$x^2 + 3x - 10 = 0$

Now, you have a quadratic equation. You can solve it by factoring or by using the quadratic formula. In this case, it's easier to factor:

$(x + 5)(x - 2) = 0$

Now, set each factor equal to zero:

$x + 5 = 0$

$x - 2 = 0$

Solve for $x$ in each equation:

For the first equation, $x + 5 = 0$, you get:

$x = -5$

For the second equation, $x - 2 = 0$, you get:

$x = 2$

So, there are two possible solutions:

$x = -5$ and $x = 2$

You should check each solution in the original equation to make sure they are valid. Logarithms are only defined for positive values, so make sure that both $x + 4$ and $x - 1$ are greater than 0 for each solution. In this case, both -5 and 2 are valid solutions.