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Ann [662]
3 years ago
13

How to solve equations in which the variable is in an exponent

Mathematics
1 answer:
Temka [501]3 years ago
3 0

Answer:

It depends on the equation.

If the bases are equal and the variables are only in the exponents, set the exponents equal.

If there are variables in the exponents, but you cannot set the bases equal, then use logarithms.

Example 1:

4^{2x + 5} = 4^9

Here you have the same base on both sides. The variables are in the exponents. Set the exponents equal and solve for x.

2x + 5 = 9

2x = 4

x = 2

Example 2:

3^{2x} = 9^{6}

The bases are different, but you can make the bases equal using laws of exponents. Remember that 9 = 3^2.

3^{2x} = (3^2)^{6}

3^{2x} = 3^{12}

Now you have equal bases, so the exponents must be equal.

2x = 12

x = 6

Example 3:

10^{x + 2} = 9

Here you can't make the bases equal, so you take the log of both sides and use laws of logs.

\log(10^{x + 2}) = \log 3^2

(x + 2) \log 10 = 2 \log 3

Recall that: \log 10 = 1

x + 2 = 2 \log 3

x = 2 \log 3 - 2

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AveGali [126]

Answer:

a) 0.25

b) 0.25

c) 0.0625

Step-by-step explanation:

The complete question is:

Do you remember when breakfast cereal companies placed prizes in boxes of cereal? Possibly you recall that when a certain prize or toy was particularly special to children, it increased their interest in trying to get that toy. How many boxes of cereal would a customer have to buy to get that toy? Companies used this strategy to sell their cereal.

One of these companies put one of the following toys in its cereal boxes: a block (B), a toy watch (W), a toy ring (R), and a toy airplane (A). A machine that placed the toy in the box was programmed to select a toy by drawing a random number of 1 to 4. If a 1 was selected, the block (or B) was placed in the box; if a 2 was selected, a watch (or W) was placed in the box; if a 3 was selected, a ring (or R) was placed in the box; and if a 4 was selected, an airplane (or A) was placed in the box. When this promotion was launched, young children were especially interested in getting the toy airplane.

What is the probability of getting an airplane in the first cereal box?

Since the machine randomly selects toys, each toy has the same probability of being obtained in a cereal box.

Then, the total outcomes are 4 and the probability of getting an airplane in the first cereal box is 0.25 (25%).

What is the probability of getting an airplane in the second cereal box?

Two independent events do not change the probability of occurrence of one event or another.

The probability of getting an airplane in the second cereal box is 0.25 (25%).

What is the probability of getting airplanes in both cereal boxes?

P(1°∩2°)= P(1°) × P(2°) = \frac{1}{4} \times \frac{1}{4} =\frac{1}{16}

P(1°∩2°)= 0.0625 = 6.25%

4 0
3 years ago
: Maria attended a Chicago Cubs baseball game last night. The ticket to get into the game was $39.00 and hot dogs cost $5.50 eac
PIT_PIT [208]

Answer: 3

Step-by-step explanation:

At first you would subtract 39 from 60. Then you would do 21/5.5 and you would get 3.818… and due to the fact you can't go over 60 you would round to 3 and not 4

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3 years ago
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yarga [219]

Answer:

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Step-by-step explanation:

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The answer would be 0.37.
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Answer:

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Step-by-step explanation:

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