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2 years ago

Find the 12th term of the geometric sequence 1, -4, 16,

Mathematics

1 answer:

Korvikt [17]2 years ago

8 0

Answer:

a₁₂ = - 4194304

Step-by-step explanation:

the nth term of a geometric sequence is

$a_{n}$ = a₁ $(r)^{n-1}$

where a₁ is the first term and r the common ratio

here a₁ = 1 and r = $\frac{-4}{1}$ = - 4 , then

a₁₂ = 1 × $(-4)^{11}$ = - 4194304

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2 years ago

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

AfilCa [17]

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

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4 months ago