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3 years ago

What is the solution of log3x − 2125 = 3

Mathematics

2 answers:

Stels [109]3 years ago

8 0

Answer:

The solution for the given expression $\log _{3x-2}\left(125\right)=3$ is $\frac{7}{3}$

Step-by-step explanation:

Given : Expression $\log _{3x-2}\left(125\right)=3$

We have to find the solution for the given expression $\log _{3x-2}\left(125\right)=3$

Consider the given expression $\log _{3x-2}\left(125\right)=3$

Apply log rule, $\log _a\left(b\right)=\frac{\ln \left(b\right)}{\ln \left(a\right)}$

$\log _{3x-2}\left(125\right)=\frac{\ln \left(125\right)}{\ln \left(3x-2\right)}$

$\frac{\ln \left(125\right)}{\ln \left(3x-2\right)}=3$

Multiply both side by $\ln \left(3x-2\right)$

We get, $\frac{\ln \left(125\right)}{\ln \left(3x-2\right)}\ln \left(3x-2\right)=3\ln \left(3x-2\right)$

Simplify , we have,

$\ln \left(125\right)=3\ln \left(3x-2\right)$

Divide both sie by 3, we get,

$\frac{3\ln \left(3x-2\right)}{3}=\frac{\ln \left(125\right)}{3}$

Also, $\frac{\ln \left(125\right)}{3}=\frac{\ln \left(5^3\right)}{3}=\frac{3\ln \left(5\right)}{3}=\ln(5)$

Thus, $\ln \left(3x-2\right)=\ln \left(5\right)$

When logs have same base, we have,

$\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\quad \Rightarrow \quad f\left(x\right)=g\left(x\right)$

Thus, $3x - 2 = 5$

Add 2 both sides, we have,

$3x = 7$

Divide both side by 3, we have,

$x=\frac{7}{3}$

Thus, the solution for the given expression $\log _{3x-2}\left(125\right)=3$ is $\frac{7}{3}$

GalinKa [24]3 years ago

3 0

The answer is x=7/3 hope this helps

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Martha wrote an example of a quadratic function for a homework assignment. The function she wrote is shown. f(x) = 5x3 + 2x2 + 7

larisa86 [58]

Answer:

The two correct options are:

The first term, 5x³, can be eliminated

The exponent on the first term, 5x³, can be changed to 2 and then combined with the second term, 2x²

Explanation:

The degree of an expression is given based on the highest power in the expression

The given expression is:

5x³ + 2x² + 7x - 3

We can note that the highest power in the given expression is 3 which means that the degree of the expression is 3

A quadratic function is a function of degree 2. This means that the highest degree in the expression (function) must be 2

Accordingly, Martha should get rid of the term with power 3 (5x³) in her equation

This can be done by 2 methods:

1- She simply eliminates the term with the third degree

2- She changes the degree of the term with degree 3 to a lower degree and combines it with any existent like term

Comparing the above with the given options, we can conclude that the two correct options are:

- The first term, 5x³, can be eliminated

- The exponent on the first term, 5x³, can be changed to 2 and then combined with the second term, 2x²

Hope this helps :)

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3 years ago

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2 years ago

Please help me find the value of x the answer must be exact

jenyasd209 [6]

Answer:

$x=11$

Step-by-step explanation:

This is a 45°-45°-90° triangle. The two legs are the same length, and the hypotenuse is that length times the square root of two:

$leg=x\\hypotenuse=x\sqrt{2}$

Therefore, the value of x is 11.

You can double-check using the sine ratio:

$sineX=\frac{opposite}{hypotenuse}$

Insert the values:

$sin45=\frac{x}{11\sqrt{2} }$

Insert the equation into a calculator:

$x=11$

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*It's important to know that, when working with trigonometry ratios, the hypotenuse is never considered the adjacent or opposite side, and the 90° angle is never used in the ratios.

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3 years ago

Can anyone help me find the coordinates (?,?) given that the shape is a parallelogram? (15 points)

xxTIMURxx [149]

The missing coordinates of the parallelogram is (m + h, n).

Solution:

Diagonals of the parallelogram bisect each other.

Solve using mid-point formula:

$$\text{Midpoint} =\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right )$

Here $x_1=m, y_1=n, x_2=h, y_2=0$

$$=\left( \frac{m+h}{2}, \frac{n+0}{2}\right )$

$$\text{Midpoint} =\left( \frac{m+h}{2}, \frac{n}{2}\right )$

To find the missing coordinate:

Let the missing coordinates by x and y.

Here $x_1=0, y_1=0, x_2=x, y_2=y$

$$\text{Midpoint}=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{x}{2}, \frac{y}{2}\right )$

Now equate the x-coordinate.

$$ \frac{m+h}{2}=\frac{x}{2}$

Multiply by 2 on both sides of the equation, we get

m + h = x

x = m + h

Now equate the y-coordinate.

$$\frac{n}{2} = \frac{y}{2}$

Multiply by 2 on both sides of the equation, we get

n = y

y = n

Hence the missing coordinates of the parallelogram is (m + h, n).

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3 years ago