What is the solution to log5(10x-1)=log5(9x=7)

1 answer:

4 0

$Domain:\\\\10x-1 > 0\ and\ 9x+7 > 0\\\\10x > 1\ and\ 9x > -7\\\\x > \dfrac{1}{10}\ and\ x > -\dfrac{7}{9}\Rightarrow x > \dfrac{1}{10}\\\\D=\left\{x|x > \dfrac{1}{10}\right\}\\\\\log_5(10x-1)=\log_5(9x+7)\iff10x-1=9x+7\qquad\text{add 1 to both sides}\\\\10x=9x+8\qquad\text{subtract 9x from both sides}\\\\x=8\in D\\\\Answer:\ \boxed{x=8}$

You might be interested in

Colin ties 5 groups of balloons to the fence. There are 3 orange balloons

maxonik [38]

Answer:

ok? what is the question?

4 0

3 years ago

Read 2 more answers

A tree that is 14 feet tall casts a shadow that is 30 feet long. Find the angle of the elevation from the top of the shadow to t

Umnica [9.8K]

Let x = angle we must find

tan x = 14/30

arctan(tan x) = arctan(14/30)

x = 25.0168934781

Answer: 25 degrees

8 0

3 years ago

If Upper X overbar equals 62, Upper S equals 8, and n equals 36, and assuming that the population is normally distributed, c

marishachu [46]

Answer:

The 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=62$ represent the sample mean

$\mu$ population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size

Part a: Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=36-1=35$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,35)".And we see that $t_{\alpha/2}=2.72$