Question

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

Part B: If (5^0)^x = 1, what are the possible values of x? Explain your answer.

Answer 1

@jdoe0001 Can you help me with this one too. I can't pas my quiz and if i don't get 80% or more i can't go on.

Answer 2

\(\bf \textit{part A, multiply both sides by }\qquad \cfrac{1}{26}
\\ \quad \\
\textit{part B, multiply both sides by }\qquad \cfrac{1}{50}\)

what does that give you?

Answer 3

they both equal 1 @jdoe0001

Answer 4

is that right?

Answer 5

\(\bf (26)x=1\implies \cfrac{1}{\cancel{26}}\cdot (\cancel{26})x=1\cdot \cfrac{1}{26}\implies x=?
\\ \quad \\ \quad \\
(50)x=1\implies \cfrac{1}{\cancel{50}}\cdot (\cancel{50})x=1\cdot \cfrac{1}{50}\implies x=?\)

Answer 6

0.0384615384615385 is part A and 0.02for part B @jdoe0001

Answer 7

\(\huge \circ.\circ\)

Answer 8

notice, the number on the left cancelled out with the denominator

Answer 9

but what happened with the exponents?

Answer 10

@jdoe0001

Answer 11

@zepdrix @jdoe0001 can you guys plz help me?

Answer 12

woops, got a typo there

\(\bf (26)x=1\implies \cfrac{1}{\cancel{26}}\cdot (\cancel{26})x=1\cdot \cfrac{1}{26}\implies \cfrac{1}{1}\cdot x=\cfrac{1}{26}\implies x=\cfrac{1}{26}
\\ \quad \\ \quad \\
(50)x=1\implies \cfrac{1}{\cancel{50}}\cdot (\cancel{50})x=1\cdot \cfrac{1}{50}\implies \cfrac{1}{1}\cdot x=\cfrac{1}{50}\implies x=\cfrac{1}{50}\)

Answer 13

i know but there is expopnents and i'm going to get this wrong.

Answer 14

ohh...shoot.. is not what you originally posted though

hmm.... need to dash in secs..... you may want to repost

Answer 15

\[\large\rm (2^{6})^x=1\]Applying a rule of exponents to the left side gives us\[\large\rm 2^{6x}=1\]Rewrite the right side as a power of 2.
2 to what power = 1?

Answer 16

ok i will repost and i will tag you k @jdoe0001

Answer 17

Is it 0? @zepdrix

Answer 18

Good. Anything to the zero power is 1. 2^0=1.
Therefore we can write our equation:\[\large\rm 2^{6x}=1\]as\[\large\rm 2^{6x}=2^0\]Since the bases are the same, we can equate the exponents.

Answer 19

how did the 2 end up a 0. and isn't the x supposed to be a 0

Answer 20

@zepdrix

Answer 21

I didn't change anything about the left side.
I rewrite 1 as 2^0.

Answer 22

Therefore, equating the exponents gives us \(\large\rm 6x=0\)
which leads to \(\large\rm x=0\)

Answer 23

For the other problem:\[\large\rm (5^0)^x = \color{orangered}{1}\]again, rewrite 1 as a power of 5,\[\large\rm (5^0)^x = \color{orangered}{5^0}\]Apply exponent rule to the left side,\[\large\rm (5^{0x}) = \color{orangered}{5^0}\]Which leads to\[\large\rm 0x=0\]And it looks like ANY value of x is going to solve this one.