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aliya0001 [1]
11 months ago
8

Log(x⁴+3x³) - log(X + 3 ) + log2 - log6 = 2logx . find the value of x

Mathematics
1 answer:
AfilCa [17]11 months ago
7 0

The given equation is

\begin{gathered} \log (x^4+3x^3)-\log (x+3)+\log 2-\log 6=2\log x \\ \log (\frac{x^4+3x^{3^{}}}{x+3})+\log \frac{2}{6}=\log x^2 \\ \log \frac{x^3(x+3)}{x+3}+\log \frac{1}{3}=\log x^2 \\ \log x^3+\log \frac{1}{3}=\log x^2 \\ \log \frac{x^3}{3}=\log x^2 \\ \frac{x^3}{3}=x^2 \\ x^3-3x^2=0 \\ x^2(x-3)=0 \end{gathered}

hence

x=0\text{ or x=3}

But x cannot be zero so x=3

So the value of x is 3

h

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An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
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DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

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At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

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Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

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Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

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Read 2 more answers
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