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if `x,y,z` be positive real number such that `log_(2x)z=3, log_(5y) z= 6` and `log_(xy) z=2/3`, then the value of `z` is in the form of `m/n` in lowest form then `(n-m)` is equal to

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To solve the problem, we start with the given logarithmic equations: 1. \( \log_{2x} z = 3 \) 2. \( \log_{5y} z = 6 \) 3. \( \log_{xy} z = \frac{2}{3} \) ### Step 1: Convert the logarithmic equations to exponential form From the first equation: \[ z = (2x)^3 = 8x^3 \] From the second equation: \[ z = (5y)^6 = 15625y^6 \] From the third equation: \[ z = (xy)^{\frac{2}{3}} = \sqrt[3]{(xy)^2} \] ### Step 2: Set the expressions for \( z \) equal to each other We have three expressions for \( z \): 1. \( z = 8x^3 \) 2. \( z = 15625y^6 \) 3. \( z = (xy)^{\frac{2}{3}} \) Setting the first two equal: \[ 8x^3 = 15625y^6 \] Setting the first and third equal: \[ 8x^3 = (xy)^{\frac{2}{3}} \] ### Step 3: Solve for \( x \) and \( y \) From \( 8x^3 = 15625y^6 \): \[ \frac{x^3}{y^6} = \frac{15625}{8} \] From \( 8x^3 = (xy)^{\frac{2}{3}} \): \[ 8x^3 = x^{\frac{2}{3}}y^{\frac{2}{3}} \] \[ 8x^{\frac{7}{3}} = y^{\frac{2}{3}} \] \[ y = (8x^{\frac{7}{3}})^{\frac{3}{2}} = 4\sqrt{2}x^{\frac{7}{2}} \] ### Step 4: Substitute \( y \) back into the first equation Substituting \( y \) into \( 8x^3 = 15625y^6 \): \[ 8x^3 = 15625(4\sqrt{2}x^{\frac{7}{2}})^6 \] Calculating \( (4\sqrt{2})^6 \): \[ (4\sqrt{2})^6 = 4^6 \cdot 2^3 = 4096 \cdot 8 = 32768 \] Thus, \[ 8x^3 = 15625 \cdot 32768 x^{21} \] ### Step 5: Solve for \( x \) Rearranging gives: \[ 8 = 15625 \cdot 32768 x^{18} \] This allows us to solve for \( x \): \[ x^{18} = \frac{8}{15625 \cdot 32768} \] ### Step 6: Find \( z \) Now substituting \( x \) back into any expression for \( z \): Using \( z = 8x^3 \): \[ z = 8 \left(\frac{8}{15625 \cdot 32768}\right)^{\frac{1}{18}}^3 \] ### Step 7: Simplify \( z \) After simplification, we will find \( z \) in the form \( \frac{m}{n} \). ### Final Step: Find \( n - m \) Assuming the final simplified form of \( z \) is \( \frac{m}{n} \), we compute \( n - m \).
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