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3 years ago

Solve 4^(x+2) = 12 for x using the change of base formula log base b of y equals log y over log b.

Mathematics

2 answers:

nalin [4]3 years ago

8 0

Answer:

x=ln(12)/ln(4)-2

Step-by-step explanation:

4^(x+2)=12

x+2=ln(12)/ln(4)

x=ln(12)/ln(4)-2

Lelu [443]3 years ago

7 0

Answer:

$x \approx -0.208$

Step-by-step explanation:

By applying logarithms to each side of the equation and some algebraic handling:

$x+2 = \log_{4} 12$

$x + 2 = \frac{\log_{10} 12}{\log_{10} 4}$

$x = \frac{\log_{10}12}{\log_{10}4} - 2$

$x \approx -0.208$

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The quantity y varies directly as x and inversely as z. When x is 10 and z is 4, y is 15. What is y when x is 20 and z is 6?

frosja888 [35]

Y=(kx)/z

15=(k10)/4
15=(5k/2)
15=k(5/2)
ties both sides by 2/5 to clear fraction
6=k

y=(6x)/z

x=20
z=6
y=(6*20)/6
y=20

8 0

3 years ago

What are the relative frequencies, to the nearest hundredth, of the rows of the two-way table?

sveta [45]

Answer: A B

Group 1 0.25 0.75

Group 2 0.44 0.56

Step-by-step explanation:

Since we have given that

Number of people of A in group 1 = 15

Number of people of B in group 1 = 45

Total number of people in group 1 is given by

$15+45=60$

Relative frequency of people of A in Group 1 is given by

$\frac{15}{60}=0.25$

Relative frequency of people of B in Group 1 is given by

$\frac{45}{60}=0.75$

Similarly, Number of people of A in group 2 = 20

Number of people of B in group 2 = 25

Total number of people in group 2 is given by

$20+25=45$

Relative frequency of people of A in Group 2 is given by

$\frac{20}{45}=0.44$

Relative frequency of people of B in Group 2 is given by

$\frac{25}{45}=0.56$

Hence, A B

Group 1 0.25 0.75

Group 2 0.44 0.56

6 0

3 years ago

This is worth 20 points

Ann [662]

45 minutes because it has 4 dots and that’s greater than the rest on the dot plot.

7 0

3 years ago

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Fraction 6/22 simplest form

asambeis [7]

Answer:

1/4

Step-by-step explanation:

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3 years ago

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Which is the solution set to the inequality (4x-3)(2x-1)&gt=0?

Serjik [45]

(4x - 3)(2x - 1) ≥ 0

First, find the zeros:

4x - 3 = 0 2x - 1 = 0

x = $\frac{3}{4}$ x = $\frac{1}{2}$

Next, plot these points and choose test points on the outside and between the zeros:

←-------0------ $\frac{1}{2}$ ------ $\frac{5}{8}$ ------ $\frac{3}{4}$ ------1------→

Lastly, plug in the test points and look for a positive result (since it is greater than 0).

Test Point 0: [4(0) - 3][2(0) - 1] = ( - )( - ) = + THIS WORKS!

Test Point $\frac{5}{8}$ : [4( $\frac{5}{8}$ ) - 3][2( $\frac{5}{8}$ ) - 1] = ( - )( + ) = - <em>This does NOT work</em>

Test Point 1: [4(1) - 3][2(1) - 1] = ( + )( + ) = + THIS WORKS!

Answer: x ≤ $\frac{1}{2}$ or x ≥ $\frac{3}{4}$

Interval Notation: (-∞, $\frac{1}{2}$ ] U [ $\frac{3}{4}$ , ∞)

Graph: ←------ $\frac{1}{2}$ $\frac{3}{4}$ --------→

5 0

3 years ago