solve the equation
log lower 6(x+9)-log lower 36 x =1
the 6 and 36 should be dropped down, then (x+9) and the x are higher.
Please show work
All your question I have answered for you are based on the rules of logs.
I think it is time that you learned these rules, and then apply them to your numerous questions
To solve the equation log₆(x+9) - log₃₆x = 1, we can use logarithm properties to simplify the equation and then solve for x.
Logarithm Property 1: logₐb - logₐc = logₐ(b/c)
Using this property, we can simplify the equation as follows:
log₆(x+9) - log₃₆x = 1
log₆(x+9) - log₆x² = 1 (since 36 = 6^2)
log₆((x+9)/x²) = 1
Logarithm Property 2: logₐb = c can be rewritten as a^c = b
Using this property, we can convert from logarithmic form to exponential form:
6^1 = (x+9)/x²
6 = (x+9)/x²
Now, we can multiply both sides of the equation by x² to eliminate the denominator:
6x² = x + 9
Rearranging the equation:
6x² - x - 9 = 0
Since this is a quadratic equation, we can factor or use the quadratic formula to find the solutions.
Factoring can be a bit challenging in this case, so let's use the quadratic formula:
The quadratic formula is given by:
x = (-b ± √(b² - 4ac))/(2a)
For our equation, the coefficients are:
a = 6, b = -1, c = -9
Using these values in the quadratic formula, we get:
x = (-(-1) ± √((-1)² - 4(6)(-9)))/(2(6))
x = (1 ± √(1 + 216))/12
x = (1 ± √217)/12
Therefore, the two possible solutions for x are:
x = (1 + √217)/12 and x = (1 - √217)/12