Hey, I would love if someone could please look over my answers. I just learned this yesterday and want to make sure I did my hw correct. Thanks!!!

Solve for x. Round answers to four decimal places where appropriate:

6. ln(5x-3)=0 ans= x=3/5
7. log5^x=3 ans= x^3=5
8. log2^32=x ans= x=16
9. log2(3x+1)=4 ans= x=7/3

ln(5x-3) = 0

5x-3 = 1
x = 3/5

Your notation is a bit odd. I think you mean

log5(x) = 3
x = 5^3 = 125

log2(32) = x
5 = x
because 32 = 2^5

log2(3x+1) = 4
3x+1 = 2^4 = 16
x = 5

^ is usually reserved for exponents, not used in logs.

Thank you again for your help! I thought it was seeming a little too easy for me to be doing it correctly.

Sure, I'd be happy to help you review your answers!

6. To solve ln(5x-3) = 0, we can start by using the inverse property of natural logarithms. Remember that ln(x) = y is equivalent to e^y = x, where e is the base of the natural logarithm. In this case, ln(5x-3) = 0 becomes e^0 = 5x-3. Since e^0 equals 1, we get 1 = 5x-3. Adding 3 to both sides gives 4 = 5x, and then dividing both sides by 5 gives x = 4/5 or 0.8.

7. To solve log5^x = 3, we can rewrite it as 5^3 = x. Using the property of logarithms, loga(b^c) = c x loga(b), we have 3 x log5(5) = log5(x), which simplifies to log5(x) = 3. Since 5^3 equals 125, we have x = 125.

8. To solve log2^32 = x, we can rewrite it as 2^x = 32. Taking the logarithm base 2 of both sides, we have x = log2(32). Evaluating this logarithm, we find that x = 5.

9. To solve log2(3x+1) = 4, we can rewrite it as 2^4 = 3x+1. Simplifying, we have 16 = 3x+1. Subtracting 1 from both sides gives 15 = 3x, and then dividing both sides by 3 gives x = 15/3 or 5.

Based on the explanations above, it seems like there are a couple of mistakes in your answers. Here are the correct answers:
6. x = 0.8
7. x = 125
8. x = 5
9. x = 5

Please double-check your calculations and make sure you understand the steps involved in solving these equations. Let me know if you have any further questions!