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2 years ago

PLEASE HELP ASAP!!! ):

Mathematics

1 answer:

stiv31 [10]2 years ago

4 0

9514 1404 393

Answer:

see attached

Step-by-step explanation:

This is made slightly more difficult by the fact that the given polynomials are not in order of decreasing powers of x. So, you need to be careful what you're adding, to make sure the terms are "like".

Add: (-2x² +3x +4) +(x² -4x -5) = (-2+1)x² +(3-4)x +(4-5) = -x² -x -1

Opp: -(3x² -x +1) = 3x² +x -1

Sub: (x² +3x -4) -(-2x² +4x -3) = (1-(-2))x² +(3-4)x +(-4-(-3)) = 3x² -x -1

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

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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

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