Logarithmic and exponential form, solving log equations, and graphing logs

I am not looking for answers, I have a test tomorrow. Any help with explaining how and if I did the processes correctly would be greatly appreciated.

Write each of the log equations in exponential form.
A) log (base 3) 81 = 4
B) log ( base B) Y = A
My answers
A) 4^3 = 81
B) A^B = Y

Write each of the exponential equations in logarithmic form.
A) 2.24 ^72 = x
B) Pi^2= 9.8696
My answers
A) log (base 72) x = 2.24
B) log (base 2) 9.8696 = pi


I mean i thought it was simple but if I use the base change formula = ln(9.8696)/ln(2)= 3.30
And pi is 3.14, is it just a litlte off or did i do it all wrong?


9. ln (x+1) - ln x = 2
ln (x+1/x) = 2
e both sides
(x+1/x) = e2
x+1 = e2 * x
??
I don't get it from here..
but the final answer is x = 1 / e^2-1

And last, graph the function
f(x) = log (base 2) 8x
ln 8x / ln 2 = ln 4x
So it would intersect at (.25, 0) rather than (1,0)?
But I don't understand where to start on the y axis...
ln(x) starts at (0. -4.5)
ln(4x) starts at (0, -3.8)
Why is that so?

Also, log (x^2)
Can you just make an x/y chart? For both of these?

log_3 (81) = 4
3^4 = 81

log_b (y) = a
b^a = y

You have the bases and exponents mixed up. With the first one, you have to make both sides of the equation powers of 3:
3^(LHS) = 3^(RHS)
A rule with logs is that if you take a base to the power of a log_(samebase), you're left with the number after the log, in this case it's 81. Besides, 4^3 is 64, not 81.
A nice way to remember how it's written is taking the [base of the log] to the power of the [right hand side of the equation] and set it equal to the [number that you're taking the log of].
--------------------------------------…
2.24^72 = x
log_2.24 (x) = 72

π^2 = 9.8696
log_π (9.8696) = 2

Same problem here, refer to the explanation above.