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3 years ago

What is the solution to the equation below?3log_4x=log_43 2+log_4 2

1 answer:

6 0

$\frac{3\log _4\left(x\right)}{3}=\frac{\log _4\left(32\right)}{3}+\frac{\log _4\left(2\right)}{3}$
Simplify
$\log _4\left(x\right)=\frac{\log _4\left(32\right)+\log _4\left(2\right)}{3}$
$\log _4\left(x\right)=1$
Apply log rule
$\log _4\left(x\right)=\log _4\left(4\right)$
Answer:
$x=4$

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elixir [45]

Answer:

Step-by-step explanation:

time taken to reach the max height is expressed according to the projectile equation;

tmax = u/g

Given u = 30m/s

g is the acceleration due to gravity = 9.81

t = 30/9.81

t = 3.0secs

hence it will take the flare 3.06secs to reach its maximum height.

Max height = u²/2g

Max height = 30²/2(9.81)

Max height = 900/19.62

Max height = 45.87m

If the people set off an emergency flare from a height of 2 meters above the water, the total height will be 45.87 + 2 = 47.87m

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3 years ago

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3 years ago

Anyone know how to do this

svlad2 [7]

Answer:

The area of the rectangle on the left side is

$9cm \: \times 4cm = 36 {cm}^{2}$

The area of the bottom rectangle is

$6cm \times 2cm = 12 {cm}^{2}$

The total area of the composite figure will be

$36 {cm}^{2} + 12 {cm}^{2} = 48 {cm}^{2}$

Step-by-step explanation:

The area of any given rectangle can be found by multiplying the length of that rectangle by its width. The rectangle on the left side has a length of 9cm but the width is unknown. To find the width, we subtract 6cm from the width of the bottom rectangle: 10cm. And that gives us 4cm.

Therefore, we can now calculate the area to be: length × width = 9cm × 4cm = 36cm²//

The area of the bottom rectangle can be found similarly by multiplying the length: 2cm by the width: 6cm of that rectangle. And the result gives us: 2cm × 6cm = 12cm²//

The total area of the composite figure is calculated by adding the results from the left and bottom rectangles together. And that gives us: 36cm² + 12cm² = 48cm²//

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3 years ago

Find the value of cos C rounded to the nearest hundredth, if necessary

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$\bold{\frac{5}{13} or 0.38}$