Mathematics
7 Online
OpenStudy (anonymous):
sqrt5^8n=125^(n+4)
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OpenStudy (anonymous):
the 8n is not under the sqrt
Parth (parthkohli):
\(\Large \color{Black}{\Rightarrow(5^{1 \over 2})^{8n} = (5^3)^{n + 4} }\)
Multiply the exponents and then just equate the exponents.
OpenStudy (lgbasallote):
\[\large 5^{\frac{1}{2}(8n)} = 5^{3(n+4)}\]
OpenStudy (cwtan):
Comparing the powers
OpenStudy (callisto):
\[\sqrt5^{8n}=125^{n+4}\]\[(5^{\frac{1}{2}})^{8n}=(5^3)^{n+4}\]4n = 3(n+4)
Can you solve it here?
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OpenStudy (lgbasallote):
\[\large 5^{4n} = 5^{3n + 12}\]
OpenStudy (anonymous):
there is no 1/2
OpenStudy (cwtan):
sqrt is power 1/2
OpenStudy (lgbasallote):
\[\sqrt 5 = 5^{\frac{1}{2}}\]
Parth (parthkohli):
\(\Large \color{Black}{\Rightarrow \sqrt[x]{y} = y^{1 \over x} }\)
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OpenStudy (cwtan):
Why @ParthKohli equation seem bigger? lol
OpenStudy (lgbasallote):
got it @zackwashere ?
Parth (parthkohli):
@cwtan latex latex latex
OpenStudy (lgbasallote):
\[\LARGE \text{like this??}\]
OpenStudy (anonymous):
nono i'm confused
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OpenStudy (cwtan):
wow pro latex
OpenStudy (lgbasallote):
where @zackwashere ?
OpenStudy (anonymous):
this is what it tells me to do....
Parth (parthkohli):
\(\huge \color{orange}{\text{HUGE!}}\)
OpenStudy (anonymous):
Which of the following is the solution to the equation
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Parth (parthkohli):
4n = 3n + 12
Solve now!
OpenStudy (anonymous):
n = 12
n = 4
n = -4
n = -12
are the answers
OpenStudy (cwtan):
solve the equation and you will get the answer
4n=3n+12
Parth (parthkohli):
Subtract 3n from both sides to isolate n.
OpenStudy (anonymous):
12 is the answer?
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Parth (parthkohli):
Yes!
OpenStudy (cwtan):
Believe urself~
OpenStudy (anonymous):
thank you.
Parth (parthkohli):
Lol yw
OpenStudy (cwtan):
Sometime Math need courage to accept the answer made by u own