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1 year ago

I need help on this logarithmic equation I need it solved step by step * It asks to solve it in terms of x. *

Mathematics

1 answer:

Rashid [163]1 year ago

3 0

To solve the function in terms of x:

$\log _{4^x}2^a=3$

find the power of the subscripts of the log

$\begin{gathered} \log _{4^x}2^a=3 \\ \log _{2^{2x}}2^a=3 \\ \frac{a}{2x}\log _22=3 \\ \text{where }\log _22=1 \\ \frac{a}{2x}=3 \end{gathered}$

cross multiply the expression

$\begin{gathered} \frac{a}{2x}=3 \\ a=3(2x) \\ a=6x \end{gathered}$

Therefore the correct answer of a = 6x

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How do you simplify this and write as a polynomial in standar form? (x-9)3x-7)+(3x^2-5x+2)

nasty-shy [4]

Answer:

Standard form of $(x-9)(3x-7) + (3x^{2} - 5x+2)$ $=6x^{2} - 39x + 65$

Step-by-step explanation:

Here, the given expression is $(x-9)(3x-7) + (3x^{2} - 5x+2)$

Now, simplifying the above expression in parts, we get

$(x-9)(3x-7) = 3x^{2} - 7x -27x + 63 = 3x^{2} - 34x + 63$

hence, combining both parts:

$(x-9)(3x-7) + (3x^{2} - 5x+2)$ $=(3x^{2} -34x +63) + (3x^{2} - 5x+2)$

= $6x^{2} - 39x + 65$

The above expression is of the STANDARD FORM: $ax^{2} +bx + c$

Hence, the standard form of $(x-9)(3x-7) + (3x^{2} - 5x+2)$ $=6x^{2} - 39x + 65$

6 0

4 years ago

A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters

In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

$\Rightarrow y=\sqrt{x^2+100}$

Differentiating with respect to t

$\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}$

Since the car driving towards the intersection at 13 m/s.

so, $\frac{dx}{dt}=-13$

$\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)$

Now

$\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)$

$=\frac{24\times (-13)}{\sqrt{676}}$

$=\frac{24\times (-13)}{26}$

= -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0

3 years ago

log8x=2 solve for x by converting the logarithm to an exponential equation. (i already know the answer is 12.5 but i need the wr

swat32

Answer:

pooooooo

Step-by-step explanation:

3 0

3 years ago

My brain is destroying itself so i need a little help I give brain award

klasskru [66]

The third one is the right one on e2020

7 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago