i have 2 math questions i don't understand so if you could help me i would like that.

Log base 3 64-log base 3 (8/3)+log base 3 2=log base 3 4r

Log base 6 (b^2+2)+logbase6 2=2

Rewrite the first as

Log(3) [64*2/(8/3)] = Log(3) 4r
If the logs to the same base of two numbers are the same, then the numbers themselves must be the same.
Therefore 128*3/8 = 4r
48 = 4r
r = 12
================
Log(6)[(b^2+2)*2] = Log(6)36
See what I did? I rewrote 2 as log(6)36
2 (b^2 + 2) = 36
b^2 + 2 = 18
Take it from there. There are two answers.

Hey I think u got the 2nd question wrong drwls...wudnt it be

Log 3 64-Log 3 (8/3) + Log 3 2 = Log 4 r

Log 3 64 - Log 3 (16/3) = Log 4r

Log 3 (63*3/16) = Log 4r

Log 3 12 = Log 4r

12=4r

r=3

Nope, drwls is correct, I got the same result as he did

remember that multiplication and division are done in the order they come

so the left side is log3[64 ÷ 8/3 x 2]
= log3[64 x 3/8 x 2]
= log3 48

etc

but in the original question was

Log 3 64-Log 3 (8/3) + Log 3 2 = Log 4 r

so, wudnt u do the addition first and then do the subtraction ... in other words solve Log 3 (8/3) + Log 3 2 and then do the rest.....?

..but it's NOT</b? Log 3 (8/3) + Log 3 2

it is -Log 3 (8/3) + Log 3 2 or
Log 3 2 - Log 3 (8/3)
= log3 (2 ÷ 8/3)
= log3 (2 x 3/8)
= log3 (6/8)

now you have log3(64 x 6/8)
= log348

oo I c it...my bad. Thnx 4 explain reiny! ._.

sorry, did not mean to print all that in bold,

forgot to turn it off after NOT

haha its aight......shud have used instead of </b? :D

Of course, I'd be happy to help you with your math questions! Let's take them one by one:

1. Log base 3 64 - log base 3 (8/3) + log base 3 2 = log base 3 4r

To solve this equation, you can start by using logarithmic properties to simplify the equation. The properties you'll need are:

- The logarithmic property, log base X (A) - log base X (B) = log base X (A/B)
- The logarithmic property, log base X (A) + log base X (B) = log base X (A * B)

Now, let's apply these properties to simplify the left side of the equation:

log base 3 64 - log base 3 (8/3) + log base 3 2 = log base 3 (64 / (8/3)) + log base 3 2
= log base 3 (64 * (3/8)) + log base 3 2
= log base 3 (24) + log base 3 2

Now that the left side of the equation is simplified, the equation becomes:

log base 3 (24) + log base 3 2 = log base 3 4r

Next, we can use another logarithmic property, log base X (A) + log base X (B) = log base X (A * B), to combine the terms on the left side:

log base 3 (24 * 2) = log base 3 4r

Simplifying further, we have:

log base 3 (48) = log base 3 4r

Now, we can apply another logarithmic property, which states that if log base X (A) = log base X (B), then A = B. Using this property, we get:

48 = 4r

Finally, to solve for r, divide both sides of the equation by 4:

r = 12

Therefore, the solution to the equation is r = 12.

2. log base 6 (b^2 + 2) + log base 6 2 = 2

To solve this equation, let's simplify the left side of the equation first. We can use the logarithmic property, log base X (A) + log base X (B) = log base X (A * B), to combine the terms:

log base 6 ((b^2 + 2) * 2) = 2

Now, we can simplify further by distributing the 2 inside the parentheses:

log base 6 (2b^2 + 4) = 2

Next, we can use the exponential property of logarithms to rewrite the equation in exponential form:

6^2 = 2b^2 + 4

Simplifying the left side, we have:

36 = 2b^2 + 4

Now, we can isolate the term with b^2 by subtracting 4 from both sides:

32 = 2b^2

Dividing both sides of the equation by 2, we get:

16 = b^2

Taking the square root of both sides, we have:

b = ± 4

Therefore, the solutions to the equation are b = -4 and b = 4.