Question

Will fan and medal.

Part A: If (6^2)x = 1, what is the value of x?
Explain your answer.

Part B: If (6^0)x = 1, what are the possible values of x? Explain your answer.

Answer 1

Is this what you mean \[6^{2x} = 1 \]

Answer 2

@Voltage

Answer 3

yep

Answer 4

we could probably use logs familiar?

Answer 5

ok

Answer 6

@Photon336 you don't need logs for these two

Answer 7

So?

Answer 8

\[6^{2x} = 1 | 6^{2*0} = 1 \]

Answer 9

so it's this \[6^{0*x} = 1 \]

Answer 10

Ok

Answer 11

, thanks for pointing that out @agent0smith actually if you take the logs of both sides you get log(1) = 0 so you don't need to use them.

Answer 12

So then If I multiply 6^2 times 1/6^2, ill get 6^2/6^2 which is equal to 1?

Answer 13

??? Hello?

Answer 14

So this is what you're saying \[36*(\frac{ 1 }{ 6^{2} }) = \frac{ 36 }{ 36 } = 1 \]

Answer 15

Yep

Answer 16

hmm

\[6^2x = 36^{x}\]

Answer 17

I think there's only one answer to this

\[36^{x} = 1 \]

x would still be zero

Answer 18

Can you maybe kind of explain this im having trouble understanding

Answer 19

yeah no problem

Answer 20

I believe the question is based off this fact:

\[a^{x} = 1 \]

\[a^{0} =1 \]

Answer 21

that when we raise any number to the zero power we get 1

Answer 22

ok

Answer 23

okay I found an alternative explanation it's a little more indepth

Answer 24

say this is the law for dividing exponents \[\frac{ n^{x} }{ n^{y} } = n^{x-y}\]

Answer 25

ok

Answer 26

first what's

\[\frac{ 5^{3} }{ 5^{2} } = ? \]

Answer 27

Well since they both have the same base and we are dividing we would need to subtract both of the exponents so our answer should be 5^1

Answer 28

I think

Answer 29

so now' what if we had something like this

\[\frac{ 5^{3} }{ 5^{3} } = 5^{3-3} = 5^{0}\]

we know that from the left hand side we've got this

\[\frac{ 5^{3} }{ 5^{3} } = 1 \]

now putting this together from the left and right hand side is this.

\[1 = 5^{0}\]

Answer 30

I don't get it

Answer 31

look at the left hand side what is 5^3/5^3

Answer 32

Umm 5^0?

Answer 33

the argument is 1 = 5^0

Answer 34

\[so~if~we~take~any~number |n^{0} = 1 \]

Answer 35

?

Answer 36

\[\frac{ n^{1} }{ n^{1} } = n^{1-1} | n~is~any~real~number | 1 = n^{0}\]

Answer 37

ok

Answer 38

This fact should be drilled into your head. Anything to the power of 0 is 1.

Eg. \[\Large \left( 55x^3 y^2 z^9 \right)^0 = 1\]

Answer 39

ok so

Answer 40

So... hopefully that fact gets drilled into your head. If there is a zero exponent on something, then it is equal to 1.

Answer 41

ok i got the answer for part A Thanks