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creativ13 [48]
3 years ago
8

Which logarithmic equation is equivalent to the exponential equation below? ea = 35

Mathematics
2 answers:
DanielleElmas [232]3 years ago
6 0

Answer: Our required logarithmic equation would be

a=\ln (35)

Step-by-step explanation:

Since we have given that

e^a=35

We need to find the logarithmic equation that is equivalent to the above exponential equation.

So, it become.,

Taking logarithmic on both the sides, we get

\ln(e^a)=\ln(35)\\\\a\ln(e)=\ln(35)\ (\because a^m=m\ln(a))}\\\\a=\ln 35\ (\because \ln e=1)

Hence, our required logarithmic equation would be

a=\ln (35)

Cerrena [4.2K]3 years ago
5 0
<span>a = ln(35) I believe is the correct answer. If correct brainliest answer plz </span>
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60! 20 times 3 is 60
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A baseball pitcher pitches a ball at the speed of 83 mph what conversion factors can be used to find the speed of the ball in fe
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You would want to convert miles per hour into feet per hour, then you should turn feet per hour into feet per second. Alternatively, you could turn miles per hour into miles per second, and then turn miles per second into feet per second

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The area is 64. Area, as you probably know, is the length times the width of the figure. The length from A to C is approximately radical 128, and the length from C to D is approximately radical 32 (btw, the radical is the check mark with line that goes above and next to the radicand (the number on the inside of the radical)). Multiply these two to get the area, and you should end up with 64.

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3 years ago
Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite
zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt  be a gram/L

Let the amount salt in the tank at any time t be Q(t).

\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}

Incoming rate = (a g/L)×(1 L/min)

                       =a g/min

The concentration of salt in the tank at any time t is = \frac{Q(t)}{10}  g/L

Outgoing rate = (\frac{Q(t)}{10} g/L)(1 L/ min) \frac{Q(t)}{10} g/min

\frac{dQ}{dt} = a- \frac{Q(t)}{10}

\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt

Integrating both sides

\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt

\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c        [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c

\Rightarrow -log|10a-15|-2=c

Therefore ,

-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2 .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0  in equation (1) we get

- log|10a|= -log|10a-15| -2

\Rightarrow- log|10a|+log|10a-15|= -2

\Rightarrow log|\frac{10a-15}{10a}|= -2

\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}

\Rightarrow 1-\frac{15}{10a} =e^{-2}

\Rightarrow \frac{15}{10a} =1-e^{-2}

\Rightarrow \frac{3}{2a} =1-e^{-2}

\Rightarrow2a= \frac{3}{1-e^{-2}}

\Rightarrow a = 1.73

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0
3 years ago
I could use some help with this problem.
Alla [95]

Answer:

A

Step-by-step explanation:

Plug in -6 to the equation

-5 - 3(-6) > 10

Simplify for -5 + 18 = 13

13 > 10

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