%3Dlog-6(16-x%5E2).jpeg "1+log6(4-x)=log 6(16-x^2)")
Solving the Equation 1 + log6(4-x) = log6(16-x^2)
In this article, we will explore the solution to the equation 1 + log6(4-x) = log6(16-x^2). This equation involves logarithms with base 6, and we will use the properties of logarithms to solve it.
Step 1: Simplify the Equation
The first step is to simplify the equation by using the property of logarithms that states:
loga(x) + loga(y) = loga(xy)
Applying this property to the left-hand side of the equation, we get:
log6(4-x) + 1 = log6(16-x^2)
Step 2: Isolate the Logarithmic Term
Next, we will isolate the logarithmic term on the left-hand side of the equation. To do this, we will subtract 1 from both sides of the equation, resulting in:
log6(4-x) = log6(16-x^2) - 1
Step 3: Use the Definition of Logarithm
Now, we will use the definition of logarithm to rewrite the equation. The definition of logarithm states that:
loga(x) = y <=> a^y = x
Applying this definition to the equation, we get:
6^(log6(4-x)) = 6^(log6(16-x^2) - 1)
Simplifying the equation, we get:
4 - x = (16 - x^2) / 6
Step 4: Solve for x
Now, we will solve for x by multiplying both sides of the equation by 6 to eliminate the fraction:
24 - 6x = 16 - x^2
Rearranging the equation to put it in quadratic form, we get:
x^2 - 6x - 8 = 0
Factoring the quadratic equation, we get:
(x - 8)(x + 1) = 0
Solving for x, we get:
x = 8 or x = -1
Therefore, the solutions to the equation 1 + log6(4-x) = log6(16-x^2) are x = 8 and x = -1.
