NIMCET 2025 — Mathematics PYQ

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Question

NIMCET 2025 — Mathematics PYQ

NIMCET | Mathematics | 2025

If $8^{x - 1} = (\frac{1}{4})^{x}$ , then find the value of: $\frac{1}{l o g _{x + 1} 4 - l o g _{x + 1} 5} + \frac{1}{l o g _{1 - x} 4 - l o g _{1 - x} 5}$

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Explanation

The solution to this problem is broken down into two main parts: finding the value of $x$ from the exponential equation, and then substituting that value to simplify the logarithmic expression.

Step 1: Solve for $x$ from the Exponential EquationWe are given: $8^{x - 1} = (\frac{1}{4})^{x}$ Express both sides using a common base of $2$ : $8 = 2^3 \implies 8^{x-1} = (2^3)^{x-1} = 2^{3x-3}\frac{1}{4} = 4^{-1} = (2^2)^{-1} = 2^{-2} \implies \left(\frac{1}{4}\right)^x = (2^{-2})^x = 2^{-2x}$ Equating the powers: 2^{3x-3} = 2^{-2x} $S in ce t h e ba ses a r e i d e n t i c a l, t h e i r e x p o n e n t s m u s t b ee q u a l :$ 3x - 3 = -2x 3x + 2x = 3 5x = 3 x = \frac{3}{5} = 0.6 $' in math mode at position 54: \dotsNow substitute$ ̲x = \frac{3}{5}…" style="color:#cc0000">Step 2: Simplify the Logarithmic BasesNow substitute $x = \frac{3}{5}$ into the bases of the given expression:First base: $x + 1 = \frac{3}{5} + 1 = \frac{8}{5}$ Second base: $1 - x = 1 - \frac{3}{5} = \frac{2}{5}$ Step 3: Simplify the Logarithmic ExpressionLet the given expression be denoted as $E$ : E = \frac{1}{\log_{\frac{8}{5}}4 - \log_{\frac{8}{5}}5} + \frac{1}{\log_{\frac{2}{5}}4 - \log_{\frac{2}{5}}5} $' in math mode at position 44: \dotsof logarithms,$ ̲\log_b a - \log…" style="color:#cc0000">Using the quotient property of logarithms, $lo g_{b} a - lo g_{b} c = lo g_{b} (\frac{a}{c})$ : E = \frac{1}{\log_{\frac{8}{5}}\left(\frac{4}{5}\right)} + \frac{1}{\log_{\frac{2}{5}}\left(\frac{4}{5}\right)} $' in math mode at position 61: \dotshing property,$ ̲\frac{1}{\log_b…" style="color:#cc0000">Using the change of base formula / base-switching property, $\frac{1}{l o g _{b} a} = lo g_{a} b$ : E = \log_{\frac{4}{5}}\left(\frac{8}{5}\right) + \log_{\frac{4}{5}}\left(\frac{2}{5}\right) $' in math mode at position 43: \dotsof logarithms,$ ̲\log_b m + \log…" style="color:#cc0000">Using the product property of logarithms, $lo g_{b} m + lo g_{b} n = lo g_{b} (m \cdot n)$ : E = \log_{\frac{4}{5}}\left(\frac{8}{5} \times \frac{2}{5}\right) E = \log_{\frac{4}{5}}\left(\frac{16}{25}\right) $' in math mode at position 13: Notice that$ ̲\frac{16}{25} $\dots" style="color:#cc0000">Notice that$ \frac{16}{25} $i s t h es q u a r eo f$ \frac{4}{5} $: \frac{16}{25} = \left(\frac{4}{5}\right)^2 Substitutethisbackintotheexpression: E = \log_{\frac{4}{5}}\left(\frac{4}{5}\right)^2 E = 2 \log_{\frac{4}{5}}\left(\frac{4}{5}\right) Since $lo g_{b} b = 1$ : E = 2 \times 1 = 2 $$

Explanation

The solution to this problem is broken down into two main parts: finding the value of $x$ from the exponential equation, and then substituting that value to simplify the logarithmic expression.

Step 1: Solve for $x$ from the Exponential EquationWe are given: $8^{x - 1} = (\frac{1}{4})^{x}$ Express both sides using a common base of $2$ : $8 = 2^3 \implies 8^{x-1} = (2^3)^{x-1} = 2^{3x-3}\frac{1}{4} = 4^{-1} = (2^2)^{-1} = 2^{-2} \implies \left(\frac{1}{4}\right)^x = (2^{-2})^x = 2^{-2x}$ Equating the powers: 2^{3x-3} = 2^{-2x} $S in ce t h e ba ses a r e i d e n t i c a l, t h e i r e x p o n e n t s m u s t b ee q u a l :$ 3x - 3 = -2x 3x + 2x = 3 5x = 3 x = \frac{3}{5} = 0.6 $' in math mode at position 54: \dotsNow substitute$ ̲x = \frac{3}{5}…" style="color:#cc0000">Step 2: Simplify the Logarithmic BasesNow substitute $x = \frac{3}{5}$ into the bases of the given expression:First base: $x + 1 = \frac{3}{5} + 1 = \frac{8}{5}$ Second base: $1 - x = 1 - \frac{3}{5} = \frac{2}{5}$ Step 3: Simplify the Logarithmic ExpressionLet the given expression be denoted as $E$ : E = \frac{1}{\log_{\frac{8}{5}}4 - \log_{\frac{8}{5}}5} + \frac{1}{\log_{\frac{2}{5}}4 - \log_{\frac{2}{5}}5} $' in math mode at position 44: \dotsof logarithms,$ ̲\log_b a - \log…" style="color:#cc0000">Using the quotient property of logarithms, $lo g_{b} a - lo g_{b} c = lo g_{b} (\frac{a}{c})$ : E = \frac{1}{\log_{\frac{8}{5}}\left(\frac{4}{5}\right)} + \frac{1}{\log_{\frac{2}{5}}\left(\frac{4}{5}\right)} $' in math mode at position 61: \dotshing property,$ ̲\frac{1}{\log_b…" style="color:#cc0000">Using the change of base formula / base-switching property, $\frac{1}{l o g _{b} a} = lo g_{a} b$ : E = \log_{\frac{4}{5}}\left(\frac{8}{5}\right) + \log_{\frac{4}{5}}\left(\frac{2}{5}\right) $' in math mode at position 43: \dotsof logarithms,$ ̲\log_b m + \log…" style="color:#cc0000">Using the product property of logarithms, $lo g_{b} m + lo g_{b} n = lo g_{b} (m \cdot n)$ : E = \log_{\frac{4}{5}}\left(\frac{8}{5} \times \frac{2}{5}\right) E = \log_{\frac{4}{5}}\left(\frac{16}{25}\right) $' in math mode at position 13: Notice that$ ̲\frac{16}{25} $\dots" style="color:#cc0000">Notice that$ \frac{16}{25} $i s t h es q u a r eo f$ \frac{4}{5} $: \frac{16}{25} = \left(\frac{4}{5}\right)^2 Substitutethisbackintotheexpression: E = \log_{\frac{4}{5}}\left(\frac{4}{5}\right)^2 E = 2 \log_{\frac{4}{5}}\left(\frac{4}{5}\right) Since $lo g_{b} b = 1$ : E = 2 \times 1 = 2 $$

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