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Question: If $10^{\log a (\log b (\log c x))} = 1$ and $10^{(\log b (\log c (\log a x)))} = 1$ then, a is equa...

If 10loga(logb(logcx))=110^{\log a (\log b (\log c x))} = 1 and 10(logb(logc(logax)))=110^{(\log b (\log c (\log a x)))} = 1 then, a is equal to

A

ab\frac{a}{b}

B

cabc^{\frac{a}{b}}

C

ab

D

cbcc^{\frac{b}{c}}

Answer

cbcc^{\frac{b}{c}}

Explanation

Solution

Here's how to solve the problem:

  1. Given Equations:

    10loga(logb(logcx))=1and10logb(logc(logax))=1.10^{\log_a(\log_b(\log_c x))} = 1 \quad \text{and} \quad 10^{\log_b(\log_c(\log_a x))} = 1.
  2. Step 1: Use the property 10y=1    y=010^y=1 \implies y=0.

    • From the first equation:

      loga(logb(logcx))=0    logb(logcx)=1(since loga1=0).\log_a(\log_b(\log_c x)) = 0 \implies \log_b(\log_c x) = 1 \quad (\text{since } \log_a 1=0).
    • From the second equation:

      logb(logc(logax))=0    logc(logax)=1.\log_b(\log_c(\log_a x)) = 0 \implies \log_c(\log_a x) = 1.
  3. Step 2: Convert logarithmic equations to exponential form.

    • For logb(logcx)=1\log_b(\log_c x) = 1:

      logcx=b    x=cb.\log_c x = b \quad \implies \quad x = c^b.
    • For logc(logax)=1\log_c(\log_a x) = 1:

      logax=c    x=ac.\log_a x = c \quad \implies \quad x = a^c.
  4. Step 3: Equate the two expressions for xx:

    ac=cb.a^c = c^b.
  5. Step 4: Solve for aa:

    Taking the cthc^{\text{th}} root,

    a=(cb)1c=cbc.a = \left(c^b\right)^{\frac{1}{c}} = c^{\frac{b}{c}}.

    Thus, the correct answer is cbcc^{\frac{b}{c}}.