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Mathematics 24 Online
OpenStudy (anonymous):

Help solving this please!

OpenStudy (anonymous):

OpenStudy (anonymous):

so, \[(4x-3)^{3/2}\] is the same as \[(\sqrt{4x-3})^3\] and \[8\] is the same as\[2^3\] so, \[\sqrt{4x-3}=2\] can you take it from here?

OpenStudy (anonymous):

x=25/4 is this correct?

OpenStudy (anonymous):

4x-3=2^2 4x-3=4 4x=7 x=7/4

OpenStudy (anonymous):

when you bring a square root to the other side, it becomes a square, so square the 2 and take it from there

OpenStudy (anonymous):

Thank you.

OpenStudy (anonymous):

does what i wrote make sense to you?

OpenStudy (anonymous):

right, anything to the power of a 1/2 is the same as square rooting it. 'index laws'

OpenStudy (anonymous):

Yes I forgot to square the 2

OpenStudy (phoenixfire):

@Chelsea04 Just a note: \[\huge x^{n\over m}=\sqrt[m]{x^n}\] rather than your\[\large (\sqrt[m]{x})^n\]. The answer x=7/4 is still correct though.

OpenStudy (anonymous):

Yea, I'm aware, it's still the same thing

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