If log 10-1-2x-x-1-x-log 10 5-1- then x equals toA- 4B- 3C- 2D-

The correct option is D 1log_10[12^x+x 1]=x[log_105 1] Now, 2^x+x 1 gt;0⇒ x gt;0 R.H.S=x(log_105 1) =x(log_105 log_1010) =xlog_10(2^ 1) =log_10(12^x) ∵ ...

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Byju's Answer

Standard XI

Mathematics

Fundamental Laws of Logarithms

If log 10[1/2...

Question

If ${log}_{10} [\frac{1}{2^{x} + x - 1}] = x [{log}_{10} 5 - 1]$ , then $x$ equals to

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Solution

The correct option is D $1$
${log}_{10} [\frac{1}{2^{x} + x - 1}] = x [{log}_{10} 5 - 1] Now, 2^{x} + x - 1 > 0 \Rightarrow x > 0 R.H.S = x ({log}_{10} 5 - 1) = x ({log}_{10} 5 - {log}_{10} 10) = x {log}_{10} (2^{- 1}) = {log}_{10} (\frac{1}{2^{x}}) ∵ L.H.S = R.H.S ∴ \frac{1}{2^{x} + x - 1} = \frac{1}{2^{x}} \Rightarrow x - 1 = 0 \Rightarrow x = 1$

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If log10[12x+x−1]=x[log105−1], then x equals to

The correct option is D 1log10[12x+x−1]=x[log105−1]Now, 2x+x−1>0⇒x>0R.H.S=x(log105−1)=x(log105−log1010)=xlog10(2−1)=log10(12x)∵ L.H.S = R.H.S∴12x+x−1=12x⇒x−1=0⇒x=1

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