What number is the same as 10^-5

If logb2=x and logb3=y, evaluate the following in terms of x and y:

Alja [10]

$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

Solution:

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

Use the following log rules

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$

Thus the given are evaluated in terms of x and y

3 0

3 years ago

Pls answer fast my teacher is waiting!

a_sh-v [17]

Answer:

For company A, y = 24x + 42

For company B, y = 28x + 25

Step-by-step explanation:

x is the number of containers

y is the total cost

For company A, y = 24x + 42

For company B, y = 28x + 25

For the cost of both companies to be the same, then

24x + 42 = 28x + 25

28x - 24x = 42 - 25

4x = 17

x = 4.25

Mr Lycan would have to order about 4 containers so the cost would be the same from each company

4 0

3 years ago

What’s the equation of a line passing through the two points (-3,3) and (0,-1)

kirill [66]

Answer:

y = -4/3x - 1

Step-by-step explanation:

3 0

3 years ago

Find the area length=x+10 width=2x+5 explain/show your steps

LenKa [72]

2x^2 + 25x + 50 not sure about the steps

3 0

3 years ago

A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of

Eva8 [605]

Answer:

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=250000$ represent the sample mean

$s=37500$ represent the standard deviation for the sample

$n=5$ sample size

$\mu_o =210250$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:

Null hypothesis: $\mu \leq 210250$

Alternative hypothesis: $\mu > 210250$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=5-1=4$

Conclusion

Since is a one right tailed test the p value would be:

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

6 0

3 years ago