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3 years ago

√x^(log(x−1)÷log(5))=5

Mathematics

1 answer:

skad [1K]3 years ago

8 0

Answer:

x=25

Step-by-step explanation:

The actual question is shown in the image attached.

Solve for x.

sqrt(x)^(log_5(x) -1) = 5 ......................(1)

Solution:

from (1)

sqrt(x)^(log_5(x) -1) = 5 .......separate exponents, law of log a^(b-c)=a^b/a^c

sqrt(x)^log_5(x) / sqrt(x) = 5 .......... cross multiply

sqrt(x)^log_5(x) = 5sqrt(x) .............. square both sides

(sqrt(x)^log_5(x))^2 = 25x .............. modify base a^(2b) = (a^2)^b

(sqrt(x)^2)^log_5(x) = 25x

x^log_5(x) = 25x .................... take log_5 on both sides

log_5(x) * log_5(x) = log_5(5^2*x) ............... simplify RHS

log_5(x) * log_5(x) = log_5(25)+log_5(x)

log_5(x) * log_5(x) = 2+log_5(x) ........ simplify

log_5(x) ^2 -log_5(x) -2 = 0 ........... substitute y = log_5(x)

y^2 - y -2 = 0

(y-2)(y+1) = 0

y=2 or

y = -1 ................... y = log_5(x) >= 0 , y=-1 rejected

y = 2

log_5(x) = 2

raise to base of 5

5^log_5(x) = 5^2

x = 25

Check by substituting x = 25 in (1)

sqrt(x)^(log_5(x) -1)

= sqrt(25)^(log_5(25) -1)

= 5^(2-1)

= 5 equal RHS, therefore solution is correct.

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Answer:

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The quantity demanded x each month of Russo Espresso Makers is 250 when the unit price p is $138. The quantity demanded each mon

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Answer:

(a) $D(q)=\frac{-1}{25} q+148$

(b) $S(q)=\frac{1}{50}q+58$

Step-by-step explanation:

(a) For the demand equation D(q) we have

P1: 138 Q1: 250

P2: 108 Q2: 1000

We can find m, which is the slope of the demand equation,

$m=\frac{p_{2} -p_{1} }{q_{2} -q_{1} }=\frac{108-38}{1000-250} =\frac{-30}{750}=\frac{-1}{25}$

and then we find b, which is the point where the curve intersects the y axis.

We can do it by plugging one point and the slope into the line equation form:

$y=mx+b\\\\D(q)=mq+b\\\\138=\frac{-1}{25}(250) +b\\\\138=-10+b\\\\138+10=b=148$

With b: 148 and m: -1/25 we can write our demand equation D(q)

$D(q)=\frac{-1}{25} q+148$

(b) to find the supply equation S(q) we have

P1: 102 Q1: 2200

P2: 102 Q2: 700

Similarly we find m, and b

$m=\frac{p_{2} -p_{1} }{q_{2} -q_{1} }=\frac{72-102}{700-2200} =\frac{-30}{-1500}=\frac{1}{50}$

$y=mx+b\\\\D(q)=mq+b\\\\72=\frac{1}{50}(700) +b\\\\72=14+b\\\\72-14=b=58\\$

And we can write our Supply equation S(q):

$S(q)=\frac{1}{50}q+58$

(c) Now we may find the equilibrium quantity q* and the equilibrium price p* by writing D(q)=S(q), which means the demand equals the supply in equilibrium:

$D(q)=S(q)\\\\\frac{-1}{25}q+148=\frac{1}{50}q+58\\\\$

$148-58=\frac{1q}{50} +\frac{1q}{25} \\\\90= \frac{1q}{50} +\frac{2q}{50}\\\\90=\frac{3q}{50}\\ \\q=1500\\\\$

We plug 1500 as q into any equation, in this case S(q) and we get:

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Which is the equilibrium price p*.

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