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3 years ago

Find the solution for: . log(3x-1)+log2=log4+log(x+2)

2 answers:

8 0

㏒(3x - 1) + ㏒(2) = ㏒(4) + ㏒(x + 2)
㏒(2(3x - 1)) = ㏒(4(x + 2))
㏒(2(3x) - 2(1)) = ㏒(4(x) + 4(2))
㏒(6x - 2) = ㏒(4x + 8)
6x - 2 = 4x + 8
- 4x - 4x
2x - 2 = 8
+ 2 + 2
2x = 10
2 2
x = 5

egoroff_w [7]3 years ago

3 0

Answer:

x=5

Step-by-step explanation:

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4 years ago

). If a, ß are zeroes of the quadratic polynomial p(x)=kx²+4x+4 such

natta225 [31]

Answer:

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Step-by-step explanation:

Product of zeros = αβ= constant / coefficient of x^2 = 4/k

Sum of zeros =α+β = - coefficient of x / coefficient of x^2= -4/k

Given

Consider a= α and b= β

$(\alpha)^2 + (\beta)^2 = 24$

$(\alpha)^2 + (\beta)^2$ can be written as $(\alpha)^2 + 2(\alpha)(\beta) + (\beta)^2$ if we add $\pm 2 (\alpha)(\beta)$ in the above equation.

$(\alpha)^2 + 2(\alpha)(\beta) + (\beta)^2 -2(\alpha)(\beta)$

$(\alpha + \beta)^2 -2(\alpha)(\beta)$

Putting values of αβ and α+β

$(\frac {-4}{k})^2 -2( \frac {4}{k}) = 24\\\frac {16}{k^2} - \frac {8}{k} = 24\\Multiplying\,\, the \,\, equation\,\, with\,\, 8K^2\\ 2 - k= 3K^2\\3k^2-2+k=0\\or\\3k^2+k-2=0\\3k^2+3k-2k-2=0\\3k(k+1)-2(k+1)=0\\(3k-2)(k+1)=0\\3k-2=0 \,\,and\,\, k+1 =0\\k= 2/3 \,\,and\,\, k=-1$

The values of k are 2/3 and -1

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4 years ago

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