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What is the value of x + y in the following two equations?
The given equations in the simplest form can be written as:
From equation (ii),
⇒ 0.6x = 3
⇒ x = 5
And 0.4y = 4/5
⇒ y = 2
So, the sum of the values of x and y is given as 7.
We know that
Put x = log_{e}(z)
The population of microbial bacteria "Gram staminee" increases exponentially as P(t) = aebt. The population of bacteria at the beginning of experiment at 10 a.m. was 20, which grows up to 40 by 12 p.m. What will be the population of bacteria at 3 p.m.?
Initially at 10 a.m., i.e. at t = 0, population of the bacteria was 20.
At 12 p.m., t = 2 and at 3 p.m., t = 5
Now, P(0) = ae^{b(0)} = ae^{0} = a = 20 (given)
Now, at t = 2, P(t) = 40
⇒ 40 = 20e^{b(2)}
⇒ e^{2b} = 2
Taking log of both sides, we get
2b = log_{e} 2 (log e = 1)
2b = 0.693
b = = 0.346
∴ P(t) = 20e^{0.346t}
∴ P(5) = 20e^{0.346 x 5}
= 20 x 5.64
= 112.8
Hence, population of bacteria at 3 p.m. = 113
10^{x} = x50
Take log of both L.H.S. and R.H.S.
L.H.S. = log (10^{x}) = x (log 10) = x (because log 10 = 1)
R.H.S = log (x^{50}) = 50 log x
L.H.S. = R.H.S.
x = 50 log x
Case I:
If x = 1, then
1 = 50 (log 1) (because log 1 = 0)
1 = 0 (Not possible)
Case II:
If x = 10, then
10 = 50 log (10)
10 = 50 (Not possible) (because log 10 = 1)
Case III:
If x = √10, then
√10= 50 log (√10)
√10 = 25 (Not possible) (because log √10 = 1/2)
Case IV:
If x = √10, then
√10 = 50 log (100) = 50 log [(10)^{2}]
= 50 × 2 log (10) (because log a = x log a)
= 50 × 2 × 1 (because log 10 = 1)
100 = 100
L.H.S. = R.H.S.
Hence, x = 100
If P = log2^{3} and Q = log6^{9}, which of the following options is true?
log2^{2} < log2^{3} < log2^{4}
i.e. 1 < P < 2
Similarly, log6^{6} < log6^{9} < log6^{36}
1 < Q < 2
Both P and Q belong to the interval (1, 2).
We shall now compare both P and Q with the average of 1 and 2, i.e. 3/2.
Let us assume that log2 3 > 3/2
Then 3 > 2^{3/2} ⇒ 3^{2} > 2^{3}
i.e. 9 > 8. This is true
P > 3/2 is correct.
We shall also suppose that log6 9 > 3/2
i.e. 9 > 6^{3/2}
9^{2} > 6^{3}
i.e. 81 > 216 is not true
Hence Q > 3/2 is not correct.
Q < 3/2
Therefore P > Q.
f(x) + f(y) =
f(x) + f(y) = log
f(x) + f(y) = log
f(x) + f(y) = f
If 4^{6 + 12 + 18 + 24 + … + 6x = (0.0625)84}, then what is the value of x?
Thus, x = 7, 8
Since x cannot be negative value, x = 7 is the answer.
The radioactive decay of a radioactive element is given as per the exponential function A(t) = A_{0} × e^{at}, where t is halflife of that element and A_{0 }is the amount of the element present initially. In a disastrous nuclear leakage incident in Chernobyl, in 1992, the level of plutonium80 was found to be 20 times the safe level. If the halflife of plutonium80 is 20 years, how long will it take for plutonium80 level to reach the safe limits?
The radioactive decay is given as per the exponential function A(t) = A0 × e^{at}.
The halflife of plutonium80 is 20 years.
After 20 years, the amount of plutonium80 left will be half.
∴ 0.5 A_{0} = A_{0} × e^{a × 20}
0.5 = e^{a × 20}
On taking log of both sides, the equation becomes:
log_{e} (0.5) = a × 20 (log e = 1)
a = 1/20 (loge 0.5)
=  = 0.0346
As it is given, after 20 years, radioactivity is reduced to onetwentieth of the initial activity.Hence, A(t) = 1/20 A0
Substituting this in the activity equation, we get
A_{0} = A_{0} e^{0.346t}
0.05 = e^{0.346t}
On taking log of both the sides, the equation becomes:
loge (0.05) = 0.0346t
2.9957 = 0.346 t
t = = 8.66 years
Now, log_{a}31 = 1
a = 31
So, (a + 1) = 32
(a + 1)^{(1/5)} = 32^{(1/5)} = 2
log10 e + loge 10
If = x, then the given expression can be written as:
Hence, the value of the given expression is always greater than 2.
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