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2^1-log₂x = 0.01

⇒ 2 ^{log₂2 - log₂x} = 0.01 = 1/100

⇒ 2 ^log2(2/x) = 1/100 (\(\because\) log A - log B = log\(\frac{A}B\))

⇒ \(\frac2x\) = \(\frac1{100}\) (\(\because\) a^log_ax = x)

⇒ x = 2 x 100 = 200

(c) 1

In a right angled triangle with a, b as sides and c as hypotenuse, 

c² = a² + b²                     (Pythagoras’ Theorem)

Now, given expression = \(\frac{log_{c+b}\,a+log_{c-b}\,a}{2\times\,log_{c+b}\,a\times\,log_{c-b}\,a}\)

= \(\frac{1}{2}\bigg[\frac{1}{\text{log}_{c-b}\,a}+\frac{1}{\text{log}_{c+b}\,a}\bigg]\)

= \(\frac{1}{2}\big[\text{log}_a(c-b)+\text{log}_a(c+b)\big]\) = \(\frac{1}{2}\big[\text{log}_a(c-b)(c+b)\big]\)

= \(\frac{1}{2}\big[\text{log}_a(c^2-b^2)\big]\) = \(\frac{1}{2}\text{log}_a\,a^2\) = log_a  a = 1.

Let \(\cfrac{log\,a}{x+y-2z}=\cfrac{log\,b}{y+z-2x}=\cfrac{log\,c}{z+x-2y}\) = k

\(\therefore\) log a = k (x + y - 2z), log b = k (y + z - 2x),

log c = k (z + x - 2y)

We have to prove that abc = 1

i.e., to prove that log (abc) = log 1

i.e. , to prove that log a + log b + log c = 0

L.H.S.= log a + log b + log c

= k (x + y -2z) + k (y + z -2x) + k (z + x -2y)

= k (x + y -2z + y + z -2x + z + x -2y)

= 0

= R.H.S.

Let \(\cfrac{log\,x}{a}=\cfrac{log\,y}{2}=\cfrac{log\,z}{5}=k\,\)

\(\therefore\) log x = ak, log y = 2k, log z = 5k ….(i)

But x⁴ y³ z^-2 = 1

Taking log on both sides, we get

log (x⁴ .y³ .z^-2 ) = log 1

\(\therefore\) log x⁴ + log y³ + log z^-2 = 0

\(\therefore\) 4 log x + 3 log y -2 log z = 0

\(\therefore\) 4(ak) + 3(2k) - 2(5k) = 0 ….[From(i)]

\(\therefore\) 4ak + 6k - 10k = 0

\(\therefore\) 4ak = 4k

\(\therefore\) a = 1

b² = ac

Taking log on both sides, we get

log b² = log ac

\(\therefore\) 2 log b = log a + log c

\(\therefore\) log a + log c = 2 log b

log₁₀(x + 1)^1/2 + log₁₀(1 – x)^3/2 = log₁₀(1 – x²)^1/2 + log¹⁰⁰10 

⇒ (x + 1)^1/2(1 – x)^3/2 = (1 – x²)^1/2 – 100 

⇒ √(1 - x²) (1 – x – 100) = 0 ⇒ x = ±1. x = -99 

But x ≠ ± 1, -99 

∴ There is no value of x exists.

log^x_0.4 = - 3 = x = (0.4)^-3 = x= (4/10)^-3

= (2/5)^-3 = (5/2)³ = 125/8

x^b = 27log⁴₃ 

= 3^3log4₃ = 3^log(4)3₃ 

= 3^log643 = 64

(c) 1

Let each ratio = k and base = e 

⇒ log_e x = k(a² + ab + b²) 

⇒ (a – b) log_e x = k (a – b) (a² + ab + b²) 

⇒ log_e x^{a – b} = k(a³ – b³) ⇒ x^{a – b} = \(e^{k(a^3-b^3)}\) 

Similarly, y^b-c = \(e^{k(b^3-c^3)}\), z^c-a = \(e^{k(c^3-a^3)}\)

∴ x^a-b . y^b-c . z^c-a = \(e^{k(a^3-b^3)}\). \(e^{k(b^3-c^3)}\) . \(e^{k(c^3-a^3)}\)

= \(e^{k[a^3-b^3+b^3-c^3+c^3-a^3]}\) = e⁰ = 1.

(d) -1

\(x\) = log_2a  a = \(\frac{\text{log}\,a}{\text{log}\,2a}\), y = log_3a  2a = \(\frac{\text{log}\,2a}{\text{log}\,3a}\)

z = log_4a  3a = \(\frac{\text{log}\,3a}{\text{log}\,4a}\)

∴ xyz - 2yz  = \(\frac{\text{log}\,a}{\text{log}\,2a}\).\(\frac{\text{log}\,2a}{\text{log}\,3a}\).\(\frac{\text{log}\,3a}{\text{log}\,4a}\) - 2\(\frac{\text{log}\,2a}{\text{log}\,3a}\).\(\frac{\text{log}\,3a}{\text{log}\,4a}\)

= \(\frac{\text{log}\,a}{\text{log}\,4a}\) - 2\(\frac{\text{log}\,2a}{\text{log}\,4a}\) = \(\frac{\text{log}\,a-2\,\text{log}\,2a}{\text{log}\,4a}\)

= \(\frac{\text{log}\,a-\,\text{log}\,(2a^2)}{\text{log}\,4a}\) = \(\frac{\text{log}\frac{a}{4}a^2}{\text{log}\,4a}\) = \(\frac{\text{log}\,(4a)^{-1}}{\text{log}\,(4a)}\) = \(\frac{-1.\text{log}\,4a}{\text{log}\,4a}\) = -1.