Question

Given that 2^(2x+2)⋅5^(x−1)=8^x⋅5^2x , evaluate 10 power x.

Answer 1

\[\large 2^{2x}+2\cdot5^{x-1}=8^{x}\cdot5^{2x}\]

Answer 2

2^2x+2

Answer 3

x 5^x-1

Answer 4

= 8 power x X 5 power 2x

Answer 5

\[\large 2^{2x+2}+2\cdot5^{x-1}=8^{x}\cdot5^{2x}\]

Answer 6

it's not +2 it's multiply

Answer 7

without a 2

Answer 8

\[\large 2^{2x}\cdot5^{x-1}=8^{x}\cdot5^{2x}\]

Answer 9

wrong!

Answer 10

2^2x+2 Multiply 5 power x-1 = ...

Answer 11

\[\large 2^{2x}+2\cdot5^{x-1}=8^{x}\cdot5^{2x}\]
Back here, then?

Answer 12

2 power (2x+2) x 5 power (x-1) = 8 power x X 5^2x

Answer 13

\[\large 2^{2x+2}\cdot5^{x-1}=8^{x}\cdot5^{2x}\]

Answer 14

finally :P

Answer 15

Sorry, you know, it's a bit hard to understand when written plainly... now hang on, let me see if this is something I'm capable of :D

Answer 16

remember to evaluate 10 power x

Answer 17

Something tells me this isn't solvable...

Answer 18

i have the answer.

Answer 19

it's not asking you to solve , evaluate 10 to the power of x

Answer 20

\[\large \frac{2^{x+2}\cdot2^{x}\cdot5^{x}}{5}=8^{x}\cdot5^{2x}\]

Answer 21

\[\large \frac{2^{x+2}\cdot10^{x}}{5}=8^{x}\cdot5^{2x}\]

Answer 22

\[\large 10^x=2^{3x}\cdot5^{2x}\cdot5\cdot2^{-x-2}\]

Answer 23

And then evaluate, I reckon. I don't think there's a value for x that would satisfy this anyway... I'm so lost :D

Answer 24

then how?

Answer 25

answer is 4/5

Answer 26

2 power x multiply 5 power x = 4/5

Answer 27

I don't think that holds... I'm missing something, probably

Answer 28

\[2^{2x + 2} \times 5^{x-1} = 8^{x} \times 5^{2x}\]

Answer 29

the 8 power x would become 2^3x

Answer 30

and it should become 3x = 2x + 2
x = 2

Answer 31

you cant assume that way x is not equal to 2 i did that and it was a mistake.

Answer 32

But I tried x = 4/5, it didn't work, neither, try it in your calculator.

Answer 33

dude 10 to the power of x = 4/5

Answer 34

oh, so it's 10^x then... hang on...

Answer 35

Right, I honestly have no idea how to go about this :/
I am shamed :(

Answer 36

Does this involve logarithms, by any chance?

Answer 37

Ok, never mind, I've reached clarity now :D

Answer 38

\[\large 10^x=2^{3x}\cdot5^{2x}\cdot5\cdot2^{-x-2}\]\[\large 10^x=2^{2x}\cdot2^{x}\cdot5^{2x}\cdot5\cdot2^{-x-2}\]\[\large 10^x=2^{2x}\cdot5^{2x}\cdot2^{x}\cdot5\cdot2^{-x-2}\]\[\large 10^x=10^{2x}\cdot5\cdot2^{x-x-2}\]\[\large 10^x=10^{2x}\cdot5\cdot2^{-2}\]

Answer 39

And now, it's easier to see that 10^x = 4/5
It turns out I gave up too soon, better not do that again :D

Answer 40

oh i see...

Answer 41

\[\large 10^x=\frac{5}{4}\cdot10^{2x}\]\[\large 0=\frac{5}{4}\cdot10^{2x}-10^x\]\[\large 0=10^x\left(\frac{5}{4}\cdot10^{x}-1\right)\]

Answer 42

So either
\[\huge 10^x = 0 \ \ or \ \ 10^x = \left(\frac{5}{4}\right)^{-1}\]
And it can't be the former.

Answer 43

Thank you!

Answer 44

No problem