log(8x²)+log(27)^1/3 = log(4/x²) i can use the law of logs to some degree in this question but just cant get it down into a single logarithm.. please help?
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$\displaystyle \log a+\log b=\log (a\cdot b)$ $\displaystyle \log a-\log b=\log\left(\frac{a}{b}\right)$
Note 27^1/3=3 So we have log(8x^2)+log3=log(4/x^2) log(24x^2)= log (4/x^2) So 24x^2= 4/x^2 x^4 = 1/6
i can follow that up to the last part :24x^2= 4/x^2 x^4 = 1/6 can u just explain how, mainly the x^4 thats confusing me where did that come from?
I should have used a space. 24x^2=4/x^2 Multiply both sides by x^2 Then get 24x^4=4 Divide both sides by 24 Then get x^4=4/24 So x^4=1/6
i understand now thats great thank you
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