how do you solve
log (x+9) = 1 - log x
log (x+9) = log10 - logx
x+9 = 10/x
x^2 + 9x - 10 = 0
(x+10)(x-1) = 0
x = -1 or 1
But logx is undefined for x < 0, so x=1 is the only solution.
check: log(1+9) = log10 = 1 - log1 = 1-0 = 1
To solve the given equation log (x+9) = 1 - log x, we can follow a step-by-step process:
Step 1: Combine the logarithms using logarithmic properties.
Start by using the property log a - log b = log (a / b) to rewrite the equation as a single logarithm:
log (x+9) + log x = 1
Step 2: Simplify the equation further.
Using the property log a + log b = log (a * b), the equation becomes:
log[(x+9) * x] = 1
Step 3: Rewrite the equation in exponential form.
By definition, log base b of a (logb a) = c can be rewritten as b^c = a. Thus, we can rewrite the equation as:
(x+9) * x = 10^1
Simplifying, we have:
x^2 + 9x = 10
Step 4: Set the equation equal to zero.
Rearrange the equation by subtracting 10 from both sides:
x^2 + 9x - 10 = 0
Step 5: Solve the quadratic equation.
Factor the quadratic equation:
(x + 10)(x - 1) = 0
This equation has two solutions:
x + 10 = 0, which gives x = -10
x - 1 = 0, which gives x = 1
Therefore, the solutions to the equation log (x+9) = 1 - log x are x = -10 and x = 1.