Working with Logarithms?

[Dillinger4]

Am i doing this correctly? 2⁴ = 16 --> log2⁽²⁾⁴ = log2¹⁶ --> 4 = log2¹⁶

And also i thought finding the root of a number was the inverse of an exponent? 5² = 25 --> Sqr(25) = 5

[Gandalf_Grey_]

yup, i am pretty sure anyways

[bugzpodder]

The log question looks correct. by the way it should write x=log_ab --- and it says x is the power you put on a to get b.

i think both a>0 and b>0 also a<>1

And also i thought finding the root of a number was the inverse of an exponent? 52 = 25 --> Sqr(25) = 5

depends on what you mean. if you are just talking about the sqrt(x) function, it says that we take the principal square root of x (or in other words, non-negative square root of x) so even though
(-5)²=25 -->sqrt(25)=5 still

[Narcoleptic]

I don't know if this helps but these are the basic rules of logarithms; they can't be argued against.

1) If a^b = c then b = log_ac (the logarithm of c to the base a)

Denotations (these are the ones I use, but can differ from person to person):

log₁₀x = log(x)
log_ex = ln(x)

The following three apply to all logarithms, not just those with base 10:

2) log(xⁿ) = n.log(x) (that's n times log x)
3) log(xy) = log(x) + log(y)
4) log(x/y) = log(x) - log(y)

Also:

5) log_aa = 1
6) log_a1 = 0 when a is not 1

And if you want to convert one base logarithm to another:

5) log_ba = log_ca / log_cb
6) log_ba = 1 / log_ab

I hope this helps.