How do I express the square root of 2 as a power of 2?

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aisha
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To solve this exponential equation (4)(2)^5x=square root of 2
I know the common base is 2 but how do i make the square root of 2 also a common base? Do i multiply both sides by ^2? When I did this I got x=1/40 I am not sure if this is the right way to do this question can someone tell me :confused:
 
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I think you might be on the right track... could you explain what you did? I might have done it the way you were thinking but I got a different answer.
 
What I did

ms. confused said:
I think you might be on the right track... could you explain what you did? I might have done it the way you were thinking but I got a different answer.

well first I found the common base of left side and got
(2^2) (2)^5x=square root of 2
then to get rid of the square root on the right I wasnt sure what to do but I decided to multiply both sides by ^2 and got [(2^2) (2)^5x]^2=2 so the exponents were [(2)(5x)]2 = 2 and then when i further simplified I got the exponent x to =1/40
 
When you got to here,
aisha said:
[(2)(5x)]2 = 2
, what did you do with the ^2 after the 2?
 
ms. confused said:
When you got to here,, what did you do with the ^2 after the 2?

I am not sure what u are talking about. :rolleyes:
 
Write both sides as powers of sqrt(2) and then compare powers.
 
From here "[(2^2) (2)^5x]^2=2"

to here

"so the exponents were [(2)(5x)]2 = 2"

You ran into some difficulty. The first equation is [tex]\left[2^{2}2^{5x}\right]^2=2[/tex], which is fine. When you combined the terms inside the [] you multiplied the exponents when you should have added them. You should get [tex]\left[2^{2+5x}\right]^2=2[/tex]. Can you manage from here?
 
shmoe said:
From here "[(2^2) (2)^5x]^2=2"

to here

"so the exponents were [(2)(5x)]2 = 2"

You ran into some difficulty. The first equation is [tex]\left[2^{2}2^{5x}\right]^2=2[/tex], which is fine. When you combined the terms inside the [] you multiplied the exponents when you should have added them. You should get [tex]\left[2^{2+5x}\right]^2=2[/tex]. Can you manage from here?

Ok I see my mistake but not sure if my answer is right now i got x=-1/5
 
No, but I think I see where you're going wrong.
Does this help (I've made a small change) ? [tex]\left[2^{2+5x}\right]^2=2^1[/tex].
 
Hold on-remember the power of 2 outside the square brackets [].

It's always a good idea after solving one of these to test your solution back in the original equation.
 
:confused: ok this is what I am doing now the exponents are (2+5x)^2=1 this to 4+10x=1 then i get -3/10 =x

oh to check myself where do i sub in x? which part?
 
Your original equation was [tex]4*(2)^{5x} = \sqrt{2}[/tex]. If you replace the "x" in this equation with "-3/10", the equation should be true.
 
shmoe said:
Your original equation was [tex]4*(2)^{5x} = \sqrt{2}[/tex]. If you replace the "x" in this equation with "-3/10", the equation should be true.

:smile: :smile: YAYAYA
IM SOO HAPPY LOL I finally got that THANKS SO MUCH everyone ESP SHMOE