How do I solve:
log6 (x^2-5x) = 1
1/2log5(r-2)-log5 r)
log2 (x-1/x^8)
I really don't know how to do these or even start
I assume the number immediately after log is the base of the log.
log6 (x^2-5x) = 1 is equivalent to writing
6^1 = x^2-5x = 6
Rewrite as a polynomial
x^2 -5x -6 = 0
(x-6)(x+1) = 0
x = 6 or -1
Do the others the same way. Rewrite the equation containing the log term as a polynomial, and factor it.
To solve logarithmic equations, we need to use properties and rules of logarithms. Let's go through each equation step by step:
1. log6 (x^2-5x) = 1
First, we need to rewrite the equation in exponential form:
6^1 = x^2 - 5x
Simplifying further:
6 = x^2 - 5x
Rearranging the equation:
x^2 - 5x - 6 = 0
Now, we can solve this quadratic equation by factoring or using the quadratic formula. Factoring gives:
(x - 6)(x + 1) = 0
Therefore, x = 6 or x = -1.
2. 1/2log5(r-2)-log5 r)
Here, we have a combination of logarithmic expressions.
To simplify this equation, we can use the properties of logarithms:
log(a) - log(b) = log(a / b)
Using this property, we rewrite the expression:
log5((r - 2)^(1/2) / r)
Now, we need to find the values for which this expression equals zero.
3. log2 (x-1/x^8)
Here we have a logarithm with multiple terms.
Let's simplify it step by step:
log2((x-1) / x^8) = log2(x - 1) - log2(x^8)
= log2(x - 1) - 8log2(x)
This equation cannot be further simplified unless you have more information or specific values for x.
Remember, when solving equations involving logarithms, it is important to check the solutions in the original equation to ensure they are valid, since logarithms are only defined for positive values.