Can someone please help me.....I was given these problems to fix the errors! x3/2 = 27/64, so x = ¾ and ln(12 + x) = (ln 12)(ln x)
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$\displaystyle \displaystyle x^{\frac{3}{2}}= \frac{27}{64} $ $\displaystyle \displaystyle (\sqrt{x})^3= \frac{27}{64} $ $\displaystyle \displaystyle \sqrt{x}= \frac{3}{4} $ $\displaystyle \displaystyle x= \frac{9}{16} $
You have ln(12+x)=ln12+lnx Therefore ln(12+x)=ln12x if you put e to both logs you get 12+x=12x---->12=11x--->x=12/11 I would normally be more wordy and explain each step but I see they don't do that here.
Originally Posted by radams33 .....I was given these problems to fix the errors! ln(12 + x) = (ln 12)(ln x) Assuming this is an error? It should be $\displaystyle \ln a +\ln b = \ln (a\times b)$
Yes I was given both of those equations to find the errors.
Thank you everyone!
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