Solve for x
log[1/3](x^(2) + x) - log[1/3] (x^(2) - x) = -1
Steps too please.
so far I tried this and I got 3 = (x+1) / (x-1)
is that base 1/3? lets call that a.
loga(x^2+x) - loga(x^2-x)=-1
loga[(x^2+x)/(x^2-x)]=-1
take the antilog of each side, base a
(x+1)/(x-1)= a^-1
looks like you are correct.
To solve for x in the equation
log[1/3](x^(2) + x) - log[1/3](x^(2) - x) = -1,
we can simplify the equation using logarithmic properties and then proceed to solve for x.
Here are the step-by-step instructions:
Step 1: Apply the quotient rule of logarithms. According to the quotient rule, when subtracting logarithms with the same base, we can rewrite it as a single logarithm of the division of the expressions inside the logarithms.
log[1/3]((x^(2) + x)/(x^(2) - x)) = -1
Step 2: Rewrite -1 as log[1/3](1).
log[1/3]((x^(2) + x)/(x^(2) - x)) = log[1/3](1)
Step 3: Apply the definition of logarithms. Since the logarithm of the base to 1 results in 1, we have:
(x^(2) + x)/(x^(2) - x) = 1
Step 4: Multiply both sides of the equation by (x^(2) - x) to eliminate the fraction.
(x^(2) + x) = (x^(2) - x)
Step 5: Expand the equation.
x^(2) + x = x^(2) - x
Step 6: Subtract x^(2) and x from both sides of the equation.
0 = -2x
Step 7: Divide both sides of the equation by -2.
0 / -2 = x
Step 8: Simplify.
x = 0
So, the solution to the equation log[1/3](x^(2) + x) - log[1/3](x^(2) - x) = -1 is x = 0.