solving for one unknown in the power: (2√2)^1-x = 1∕8√2

[cookie101]

can someone teach me how to do this? heres a e.g question from my booklet.


(2√2)^1-x = 1∕8√2

if ur confused about the question it is (2.squareroot2)power is 1-x = 1/8.square root2

thank you please help this is due this week.

 

[Deleted member 4993]

can someone teach me how to do this? heres a e.g question from my booklet.


(2√2)^1-x = 1∕8√2

if ur confused about the question it is (2.squareroot2)power is 1-x = 1/8.square root2

thank you please help this is due this week.

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[pka]

can someone teach me how to do this? heres a e.g question from my booklet.
(2√2)^1-x = 1∕8√2

\(\displaystyle \Large 2\sqrt2=2^{\frac{3}{2}}~\&~\dfrac{\sqrt2}{8}=2^{ \frac{-5}{2}} \)

So you have \(\displaystyle \Large 2^{\frac{3-3x}{2}}=2^{ \frac{-5}{2}} \)

 

[Otis]

\(\displaystyle 2\sqrt2=2^{\frac{5}{2}}\)

That's not correct.

 

[cookie101]

That's not correct.

do u know how to do it?

 

[Deleted member 4993]

do u know how to do it?

Where is YOUR work? Please show some effort!!

(2√2)^(1-x) = 1∕8 * √2

First of all, those added () and * are very important.

2√2 = 2¹ * 2^1/2 = 2^(1+1/2) = 2^3/2

1/8 * √2 = 2^-3 * 2^1/2 = 2^-5/2

(2√2)^(1-x) = 1∕8 * √2

(2^3/2)^(1-x) = 2^-5/2

2^[3/2*(1-x)] = 2^-5/2

3/2 * (1-x) = -5/2

Now continue......