Exponential Eqn

[aisha]

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 im not sure if this is the right way to do this question can someone tell me :confused:

[futb0l]

4*(\sqrt{2} * \sqrt{2})^{5x} = \sqrt{2}

[ms. confused]

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.

[aisha]

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

[ms. confused]

When you got to here,[(2)(5x)]2 = 2, what did you do with the ^2 after the 2?

[aisha]

When you got to here,, what did you do with the ^2 after the 2?

I am not sure what u are talking about. :uhh:

[Gokul43201]

Write both sides as powers of sqrt(2) and then compare powers.

[aisha]

lol sorry still confused can someone show me :uhh:

[shmoe]

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

[aisha]

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

Ok I see my mistake but not sure if my answer is right now i got x=-1/5

[Gokul43201]

No, but I think I see where you're going wrong.
Does this help (I've made a small change) ? \left[2^{2+5x}\right]^2=2^1.

[shmoe]

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.

[aisha]

: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?

[shmoe]

Your original equation was 4*(2)^{5x} = \sqrt{2} . If you replace the "x" in this equation with "-3/10", the equation should be true.

[aisha]

Your original equation was 4*(2)^{5x} = \sqrt{2} . If you replace the "x" in this equation with "-3/10", the equation should be true.

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