how do i solve these two problems?

3log(base5)^(x^2+9)-6 = 0
log(base8)^(n-3) + log(base8)^n+4)= 1

can you tell me step by step as well as the answers? please and thank you

Here is how to do the first one. The second is similar. First simplify the algebra. Then write the log equation in a different form.

3 log(base5)^(x^2+9) = 6
log(base5)^(x^2+9) = 2
Which is another way of saying:
5^2 = 25 = x^2 + 9
x^2 = 16
x = + or - 4

how'd you get the log(base5)^(x^2+9) = 2 ?

I divided both sides of the first equation by 3. It's a perfectly "legal" operation when dealing with algebraic equations.

Sure! I can help you solve both of these problems step by step.

Problem 1: 3log(base5)^(x^2 + 9) - 6 = 0

Step 1: Move the constant term to the other side of the equation.
3log(base5)^(x^2 + 9) = 6

Step 2: Divide both sides of the equation by 3 to isolate the logarithm.
log(base5)^(x^2 + 9) = 2

Step 3: Apply the logarithmic property to the equation.
x^2 + 9 = 5^2

Step 4: Simplify the equation.
x^2 + 9 = 25

Step 5: Subtract 9 from both sides of the equation.
x^2 = 16

Step 6: Take the square root of both sides to solve for x.
x = ±4

So the solutions to the equation are x = 4 or x = -4.

Problem 2: log(base8)^(n - 3) + log(base8)^(n + 4) = 1

Step 1: Combine the two logarithms using the product property.
log(base8)^((n - 3)(n + 4)) = 1

Step 2: Rewrite the equation in exponential form.
8^1 = (n - 3)(n + 4)

Step 3: Simplify the equation.
8 = n^2 + n - 12

Step 4: Move all terms to one side of the equation.
n^2 + n - 12 - 8 = 0
n^2 + n - 20 = 0

Step 5: Factorize the quadratic expression.
(n - 4)(n + 5) = 0

Step 6: Set each factor equal to zero and solve for n.
n - 4 = 0 or n + 5 = 0
n = 4 or n = -5

So the solutions to the equation are n = 4 or n = -5.

I hope this helps! Let me know if you have any further questions.