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2 years ago

What is the solution to log Subscript 5 Baseline (10 x minus 1) = log Subscript 5 Baseline (9 x + 7)

Mathematics

2 answers:

NARA [144]2 years ago

5 0

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$\sf{log_{5}}(\tt{10x-1})=\sf{log_{5}}(\tt{9x+7})\\\\10x-1=9x+7\\\\10x-9x=7+1\\\\\pmb{\green{\underline{\sf{x=\frak{8}}}}}$

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zimovet [89]2 years ago

4 0

Answer:

x = 8

Step-by-step explanation:

Given:

log₅ (10x - 1) = log₅ (9x + 7)
10x - 1 = 9x + 7
10x - 9x = 7 + 1
x = 8

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3 years ago

A buoy floating in the sea is bobbing in simple harmonic motion with period 2 seconds and amplitude 8 in. Its displacement d fro

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The equation of the displacement $d$ as a function of time $t$ is :

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Step-by-step explanation:

Generally , A simple harmonic wave is a sinusoidal function that is it can be expressed in simple $sin$ or $cos$ terms.

Thus,

$d(t) = Asin(wt+c)$

is the general form of displacement of a SHM.

where,

$d(t)$ is the displacement with respect to the mean position at any time $t$
$A$ is amplitude
$w$ is the natural frequency of oscillation ( $rads^{-1}$ )
$c$ is the phase angle which indicates the initial position of the object in SHM ( $rad$ )

given,

Time period ( $T$ ) = $2s$
$A=8$
The natural frequency ( $w$ ) and time period ( $T$ ) is :

$w=\frac{2\pi} {T}$

∴

$w$ = $\frac{2\pi }{2} = \pi$ $rads^{-1}$

∴

the equation :

⇒ $d(t)=8sin(\pi t+c)$ ------1

since $d=0$ when $t=o$ ,

⇒ $0=8sinc\\c=n\pi$ ------2

where n is an integer ;

⇒since the bouy immediately moves in the negative direction , $x$ must be negative or c must be an odd multiple of $\pi$ .

⇒ $d(t) = 8sin(\pi t+(2k+1)\pi )$ ------3

where k is also an integer ;

the least value of $k=0$ ;

thus ,

the equation is :

$d(t)=8sin(\pi t+\pi )$

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3 years ago