Sdlve the felowng and explaun process in details,a) If logy (m-n) = 0 and logu Cantz)=2, determneandthe Values of onb) Sdwe the equatiqn 63**! 2xLeave gour answers in exact férmc) Solvle for x given 20gu(x+4)-log«(x+12)=1|2K-3a) The half life of an element can be modellod byM= Mo tselement;t is the elapsed fime, in houes; and H is HheMass that remains after Fime t. What isof the elemant ?45 where Mo is the initial mass of ine32the half-life

Answered: Save the tollowng and expratn process…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Sdlve the felowng and explaun process in details,
a) If logy (m-n) = 0 and logu Cantz)=2, determne
and
the Values of on
b) Sdwe the equatiqn 63**! 2x
Leave gour answers in exact férm
c) Solvle for x given 20gu(x+4)-log«(x+12)=1|
2K-3
a) The half life of an element can be modellod by
M= Mo ts
element;t is the elapsed fime, in houes; and H is Hhe
Mass that remains after Fime t. What is
of the elemant ?
45 where Mo is the initial mass of ine
32
the half-life

Expert Solution

Step 1

Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.

(a)

The given equations are $\log_{4} (m - n) = 0 a n d \log_{4} (m + 2) = 2$ .

Simplify the first equation as follows.

$\log_{4} (m - n) = 0 m - n = 4^{0} (I f \log_{a} x = b t h e n x = a^{b}) m - n = 1 (1)$

Simplify the second equation as follows.

$\log_{4} (m + 2) = 2 m + 2 = 4^{2} m + 2 = 16 m = 14$

Substitute m=14 in equation (1) as follows.

$m - n = 1 14 - n = 1 n = 13$

Therefore, the values of m and n are 14 and 13 respectively.

Step 2

(b)

Solve the given equation as follows.

$6^{3 x + 1} = 2^{2 x - 3} \ln (6^{3 x + 1}) = \ln (2^{2 x - 3}) (3 x + 1) \ln 6 = (2 x - 3) \ln 2 3 x \ln 6 + \ln 6 = 2 x \ln 2 - 3 \ln 2 3 x \ln 6 = 2 x \ln 2 - 3 \ln 2 - \ln 6 3 x \ln 6 = 2 x \ln 2 - (\ln 2^{3} + \ln (6)) (a \ln x = \ln x^{a}) 3 x \ln 6 - 2 x \ln 2 = - \ln 48 (\ln m + \ln n = \ln (m n)) x (3 \ln 6 - 2 \ln 2) = - \ln 48 x (\ln (\frac{216}{4})) = - \ln 48 (\ln m - \ln n = \ln (\frac{m}{n})) x = - \frac{\ln 48}{\ln 54}$

Therefore, the value of x is $x = - \frac{\ln 48}{\ln 54}$ .

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