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3 years ago

6

How do logarithmic equations relate to exponential equations?

1 answer:

mash [69]3 years ago

6 0

Logarithmic equations are the inverse of exponential equations.

Say you have the exponential equation of:

$y = 2^x$

This is inversely equivalent to:

$log_{2}(y) = x$

By general definition:

$y = b^x$ = $log_{b} (y) = x$

You might be interested in

(w^5 z^2 )(−9w^2 z^5)

Lady_Fox [76]

We have that
(w^5*z^2)*(-9*w^2*z^<span><span>5)</span>

</span><span>Multiplying exponential expressions
we know that
</span>w^5<span> multiplied by </span>w^<span>2 = w</span>^<span>(5 + 2) = w</span>^7
and
z^2<span> multiplied by </span>z^<span>5 = z</span>^<span>(2 + 5) = z</span>^7

therefore
(w^5*z^2)*(-9*w^2*z^5) =-9*( w^7)*( z^7)
-9*( w^7)*( z^7)=-(3^2)*( w^7)*( z^7)

the answer is
-(3^2)*( w^7)*( z^7

5 0

3 years ago

Meaning of bench mark in math

mezya [45]

It mean benchmarks in out side of math
a criterion by which to measure something; standard; reference point

8 0

4 years ago

Bob has 1.5 pounds of nails. If there are 80 nails in a pound, how many nails does Bob have?

amm1812

Answer:

120 nails.

Step-by-step explanation:

80 x 1.5 = 120.

80 x 1 = 80

80 x 0.5 = 40

80 + 40 = 120

3 0

3 years ago

Read 2 more answers

Consider a series system composed of 4 separate components where each component has a 30% chance of failing. Assume each compone

Marina86 [1]

Answer:

16.15% probability that exactly 3 of them would function

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability of each system working:

4 components, which means that $n = 4$

Each has a 30% probability of failing, so $p = 1 - 0.3 = 0.7$

For the system to work, all 4 components have to work. This is P(X = 4).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{4,4}.(0.7)^{4}.(0.3)^{0} = 0.2401$

0.2401 probability of a system working.

If you have 7 of these systems, what is the probability that exactly 3 of them would function?

Now 7 systems, so $n = 7$

0.2401 probability of a system working.

We have to find P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{7,3}.(0.2401)^{3}.(0.7599)^{4} = 0.1615$

16.15% probability that exactly 3 of them would function

5 0

3 years ago

Iris is building a model of the Washington Monument. Her model measures 15 in. tall and is 1.5 in. wide at the base. The actual

Sedbober [7]

Actual W/ actual height= model W/ model height
55 ft/ actual height= 1.5 in./ 15 in,
actual height=55 ft*15 in./1.5 in= 550 ft

5 0

3 years ago

Read 2 more answers