Help Solving Logarithmic Equations! What is log(base d) 12=6?

Question

log(base d)=6
I know that you have to change it to exponential form like this:
d^6=12
So what number to the 6th equals 12?

Answer 1

You need to get d all by itself, or to the power of 1 (an exponent of 1)
To do this, hit each side with a 1/6 power
d^(6 x 1/6) = 12 ^(1/6)
so you get
d^(1) = 12 ^(1/6)
Now, all you have to do is take 12 to the 1/6 power.
Any calculator made in the past 10 yrs can do that.

Answer 2

considering that 2^4 = 16 you could rewrite the equations as 2^(x - x^2) = 2^-4x remove the bases x - x^2 = -4x upload 4x 5x - x^2 = 0 aspect x(5 - x) = 0 split and clean up 5x = 0 --> x = 0.5 - x = 0 --> x = 5 you could also try this with logs take the log of both area. keep in mind that the log of a range to a ability = the flexibility situations the log (the exponent comprises the front) so (x - x^2) log 2 = x log(a million/16) divide both area by log(a million/16) (-a million/4)(x - x^2) = x multiply by -4 x - x^2 = -4x relax has similarities as above

Answer 3

I dont have a calculator but it is roughly 1.5

if d^6 = 12 then the 6th root of 12 = d

Answer 4

it's not gonna be a nice and round number, yes.

clearly, d > 1
clearly, d < 2, cause 2^6 = 62

so basically what you do is take the root of 6th degree from 12... regular windows calculator can do that.... and it tells me

1.5130857494229015887840596903103

Answer 5

In the following, log means log base d:-
log 12 = 6
12 = d^6
12^(1/6) = d
d = 12^(0.167)
Now use y^x button on calculator where y = 12 and x = 0.167:-
d = 1.51

Answer 6

d^6=12
than you take the 6th root of both sides
becuase nth root of nth power is itself
so you get d by itself that way and it approximately 1.51

Answer 7

somewhere in between 1.4 and 1.45