Question

6^5^x=20

Answer 1

like this?
\[6^{5^x}=20\]

Answer 2

yeah

Answer 3

ok,
since we are working with an exponent and the x is in the exponent, you'll want to apply the log power rule that looks like this:
\[\log(m^{n^x}) = x*\log(m^n)\]

Answer 4

do you see how you can apply that to the equation you have?

Answer 5

no can you put it together

Answer 6

if the equation given to me looked like this:
\[m^{n^x} = z\]
the I would log both sides:
\[\log(m^{n^x}) = \log(z)\]
then I would use the log power rule
\[x*\log(m^{n}) = \log(z)\]

Answer 7

do you see how
6 is in the same place as m and
5 is in the same place as n and
x is in the same place as x and
z is in the same place as 20 ?

Answer 8

\[6^{5^x}=20\]
\[m^{n^x}=z\]

Answer 9

i got -2.5897

Answer 10

hmmm....
it should be 0._____ number
can you post up your steps and I can help?

Answer 11

@DemolisionWolf Using the power log rule, wouldn't the expression be \[n^xlog(m)=\log(z)\] I don't think you can bring down the "x" from the very top like that.

Answer 12

my steps were 6^5^x=20

and then i took the log of both sides and briught the x down

and i got -2.5897

Answer 13

so once we get here:
x log(6^5) = log(20) we do
x= log(20) / log(6^5)

Answer 14

oh i didnt divide

Answer 15

@OrionsBelt ya, are right, thanks a ton!

Answer 16

i got 0.3343

Answer 17

@kavon orionsbelt was right, I slipped up, so we will have to redo a step, and do one more after it.

0.3343 is what I got, but I used the log power rule wrong.

Answer 18

start from her again
\[n^xlog(m)=\log(z)\]

Answer 19

5^x log(6) = log(20)
becomes
5^x = log(20)/log(6)

Answer 20

and then i calculate from there?

Answer 21

the x is still in the exponent spot, so we have to get it down by doing the log power rule again, so we log both sides again, then apply the log-power-reduction-rule again

Answer 22

your answer will be close to the 0.33 we got before.

Answer 23

ok