solve: log base2(log basex(x+6)=1
There's 2 different bases, how do I solve for x?
To solve the equation log base 2 of log base x of (x+6) equals 1, we need to combine the logarithmic functions into a single equation.
Step 1: Let's rewrite the equation using the logarithmic properties.
log base 2 (log base x (x + 6)) = 1
Step 2: Convert the logarithmic equation into an exponential equation. We can rewrite it as:
2^1 = log base x (x + 6)
Simplifying further,
2 = log base x (x + 6)
Step 3: Rewrite the equation using the definition of logarithms. In logarithmic form, we have:
x + 6 = x^2
Step 4: Rearrange the equation to set it equal to zero. Let's rewrite it as:
x^2 - x - 6 = 0
Step 5: Factorize the quadratic equation to solve for x. Factoring the quadratic equation, we get:
(x - 3)(x + 2) = 0
Step 6: Set each factor equal to zero and solve for x. We have two cases:
Case 1: x - 3 = 0
x = 3
Case 2: x + 2 = 0
x = -2
Therefore, the values of x that satisfy the equation are x = 3 and x = -2.