All categories

3 years ago

What is the value of x for log6(5x) + log6(2)=5

1 answer:

8 0

$\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log_6(5x)+\log_6(2)=5\implies \log_6(5x\cdot 2)=5\implies \log_6(10x)=5 \\\\\\ 6^{ \log_6(10x)}=6^5\implies 10x=6^5\implies x = \cfrac{6^5}{10}\implies x = \cfrac{3888}{5}\implies x = 777.6$

You might be interested in

Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1

Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

$P(x=r)=^nC_rp^rq^{n-r}$

(a) Exactly two will be drunken drivers.

$P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}$

$P(x=2)=66(0.2)^{2}(0.8)^{10}$

$P(x=2)=\approx 0.28347$

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

$P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)$

$P(x=3\text{ or }x=4)=P(x=3)+P(x=4)$

Using binomial we get

$P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}$

$P(x=3\text{ or }x=4)=0.236223+0.132876$

$P(x=3\text{ or }x=4)\approx 0.369099$

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

$P(x\geq 7)=1-P(x$

$P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]$

$P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]$

$P(x\leq 7)=1-[0.9961]$

$P(x\leq 7)=0.0039$

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

$P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)$

$P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315$

$P(x\leq 5)=0.9806$

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

5 0

3 years ago

If the world's population increased exponentially from billion in 1998 to billion in 2008 and continued to increase at the same

Gelneren [198K]

Answer:

The answer is "7.122 billion"

Step-by-step explanation:

Please find the complete question in the attached file.

Using formula:

$\to y=Ce^{kt}$

$C=5.937\\\\y=6.771\\\\t=10\\\\ k=?6.771=5.937e^{10k}\\\\1.1405=e^{10k}\\\\\ln(1.1405)={10\ k}\\\\0.013={k}\\\\$

So,

$\to y=5.937e^{.013t}\\\\$

when it becomes $2012, \ t=14$ so plug the value into the formula

$\to y= 5.937e^{.013(14)} =7.122 \ billion$

7 0

3 years ago

Please help!! Time limit!

gavmur [86]

Answer:

Step-by-step explanation:

92÷2= 46

188÷ 1/4=752

276÷ 1/6= 1656

So the third one is the answer

6 0

3 years ago

A car traveled on the highway at 65 miles per hour for 45 minutes. What distance did the car travel?

daser333 [38]

i think its 52 miles.

the fraction of the time is 4/5 which is .8 so

65 * .8 = 52

i hope this helps :)

3 0

3 years ago

On a city map 3/4 inch corresponds to 7 1/2 miles bonnie drives from her house to a movie theater and then from a movie theater

Juliette [100K]

Answer:

Option D: $12\frac{1}{2}$ is the correct answer.

Step-by-step explanation:

Given that:

3/4 inches on map are equal to $7\frac{1}{2}$ miles

We will find the unit rate in term of inches.

$\frac{3}{4}\ inches = 7\frac{1}{2}\\\\Unit\ rate = \frac{15}{2}*\frac{4}{3}\ miles\\\\Unit\ rate = 10\ miles\ per\ inch$

Total distance covered by Bonnie according to map;

Total distance = $\frac{1}{2}+\frac{3}{4} = \frac{2+3}{4}$

Total distance = $\frac{5}{4}\ miles$

Actual distance = $\frac{5}{4}*10 = \frac{25}{2}\ miles \\$

Actual distance = $12\frac{1}{2}\ miles$

Hence,

Option D: $12\frac{1}{2}$ is the correct answer.

8 0

3 years ago