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adoni [48]
1 year ago
9

Use the properties of exponential and logarithmic functions to solve each system. Check your answers.

Mathematics
1 answer:
BlackZzzverrR [31]1 year ago
3 0

The solved logarithmic function is 2x - y = e and x + y = 8 for log (2x - y) = 1 and log (x + y) = 3 log 2 respectively.

What is a logarithmic and exponential function?

Logarithmic functions and exponential functions are inverses of each other. The logarithmic function is denoted by using the word log while exponential by using the alphabet,  e. For example log 10 = 1 and e^2.

Solving the given expressions: log (2x - y) = 1 and log (x + y) = 3 log 2

Applying given properties of logarithmic and exponential function;

log a + log b = log (ab)

4 log x = log x⁴

e^(log x) = 1

e^(x + y) = (e^x) × (e^y)

Take expression, log (2x - y) = 1

Applying exponential on both sides, we get,

e^(log 2x - y) = e^1

2x - y = e

To Check Results,

Taking logarithm both sides,

log (2x - y) = log e

log (2x - y) = 1

Thus, the answer is verified.

Take expression, log (x + y) = 3 log 2

Applying exponential on both sides, we get,

e^(log (x + y) = e^(3 log 2)

x + y = e^(log 8)

x + y = 8

To Check Results,

Taking logarithm both sides,

log (x + y) = log 8

log (x + y) = 3 log 2

Thus, the answer is verified.

Hence, the solved expression is 2x - y = e and x + y = 8 for log (2x - y) = 1 and log (x + y) = 3 log 2 respectively.

To learn more about the logarithmic and exponential functions, visit here:

brainly.com/question/13473114

#SPJ4

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Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2
iren2701 [21]
Hello there, hope I can help!

I assume you mean L\left\{\frac{ekt+e-kt}{2}\right\}
With that, let's begin

\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \  L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}

\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}
\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b
L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}
\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}

L\left\{t\right\} \ \textgreater \  \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \  L\left\{t\right\}=\frac{1}{s^2}

L\left\{\frac{e}{2}\right\} \ \textgreater \  \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \  L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \  \frac{e}{2s}

\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}

\frac{ek}{2}\cdot \frac{1}{s^2}  \ \textgreater \  \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \  \frac{ek\cdot \:1}{2s^2} \ \textgreater \  \mathrm{Apply\:rule}\:1\cdot \:a=a
\frac{ek}{2s^2}

\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \  \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \  \frac{k\cdot \:1}{2s^2} \ \textgreater \  \mathrm{Apply\:rule}\:1\cdot \:a=a
\frac{k}{2s^2}

\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}

Hope this helps!
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If f(x) = 3x + 2, what is f(5)?
iVinArrow [24]

Hi there!

\large\boxed{f(5) = 17}}

f(x) = 3x + 2, find f(5):

To find f(5), we simply substitute in 5 for x:

f(5) = 3(5) + 2

f(5) = 15 + 2

f(5) = 17

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