\[\sqrt{16x ^{3}}/\sqrt{4x}\]
start by simplifying the constants. What do you notice about 4 and 16?
factor 4x
or just 4
A couple properties you should know first\[\sqrt x=x^{1/2}\]\[\sqrt ab=\sqrt a \sqrt b\]\[(x^a)^b=x^{ab}\]
start with the constants. You should see immediately that they are perfect squares
4 and 2
yes. so now you have \[4\sqrt x^3 \over 2\sqrt x\]
change the radical sign to 1/2 power to make things easier
\[4(x^{1/2})^3 \over 2x^{1/2}\]
now looking at the properties I posted finish the problem
\[{x^a \over x^b }= x^{a-b}\]
x^1/2(3) for the top half right
keep going
4(x)^1/3
No i would multiply the bottom half now
\[\color{red}4(x^{1/2})^3 \over \color{red}2x^{1/2}\]\[\color{red}2(x^\color{blue}{{1/2}})\color{blue}{^3} \over x^{1/2}\]\[\color{red}2x^\color{blue}{{3/2}} \over x^{1/2}\]Now simplify the numerator and denominator using:\[{x^a \over x^b }= x^{a-b}\]
ok so it would be 2x(3/2-1/2= 2/2 which is one ) so just 2x
correct!! Good job! Do you understand all the steps?
yea i just need to remember the rules
log\[\log_{6} (x-5)+\log_{6} ^x=2\]
\[\log_a x + \log_a y=\log_a xy\]\[\log_a x - \log_a y = \log_a x/y\]
\[\log_a x = y \rightarrow a^y = x\]
you should be able to solve now. take your time, I'm just making dinner
BTW, once you get rid of the log, you should see a polynomial. You need to get it in the form: 0 = ax^2 + bx +c, so that you can factor or use the quadratic to solve for x
what do i do with the 2
see my 3rd equation
\[\log_{6} (x-5)+\log_{6} x=2\]\[\log_6 x(x-5)=2\]\[\log_6 x^2-5x=2\]\[x^2-5x=6^2\]\[x^2-5x-36=0\]You should be at this point. The log is gone and we have to solve for x. We can factor it or use the quadratic. Are you familiar with either of those methods?
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