If [45]−3×[45]−5=[45]3x−2\Big[\dfrac{4}{5}\Big]^{-3} \times \Big[\dfrac{4}{5}\Big]^{-5} = \Big[\dfrac{4}{5}\Big]^{3x - 2}[54]−3×[54]−5=[54]3x−2, the value of x is:
2
12\dfrac{1}{2}21
-2
−12-\dfrac{1}{2}−21
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According to product property,
am×an=am+na^m \times a^n = a^{m + n}am×an=am+n
[45]−3×[45]−5=[45]3x−2⇒[45]−3+(−5)=[45]3x−2⇒[45]−8=[45]3x−2\Big[\dfrac{4}{5}\Big]^{-3} \times \Big[\dfrac{4}{5}\Big]^{-5} = \Big[\dfrac{4}{5}\Big]^{3x - 2}\\[1em] ⇒ \Big[\dfrac{4}{5}\Big]^{-3 + (-5)} = \Big[\dfrac{4}{5}\Big]^{3x - 2}\\[1em] ⇒ \Big[\dfrac{4}{5}\Big]^{-8} = \Big[\dfrac{4}{5}\Big]^{3x - 2}\\[1em][54]−3×[54]−5=[54]3x−2⇒[54]−3+(−5)=[54]3x−2⇒[54]−8=[54]3x−2
Using am=an⇒m=na^m = a^n \Rightarrow m = nam=an⇒m=n
⇒−8=3x−2⇒−8+2=3x⇒−6=3x⇒x=−63⇒x=−2\Rightarrow -8 = 3x - 2\\[1em] \Rightarrow -8 + 2 = 3x\\[1em] \Rightarrow -6 = 3x\\[1em] \Rightarrow x = \dfrac{-6}{3}\\[1em] \Rightarrow x = -2⇒−8=3x−2⇒−8+2=3x⇒−6=3x⇒x=3−6⇒x=−2
Hence, option 3 is the correct option.
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If x=3mx = 3^mx=3m and y=3m+2,xyy = 3^{m + 2}, \dfrac{x}{y}y=3m+2,yx is:
9
19\dfrac{1}{9}91
6
[x−23y−1]−1\Big[\dfrac{x^{-2}}{3y^{-1}}\Big]^{-1}[3y−1x−2]−1 is equal to:
3x2y\dfrac{3x^2}{y}y3x2
x23y\dfrac{x^2}{3y}3yx2
y3x2\dfrac{y}{3x^2}3x2y
3yx2\dfrac{3y}{x^2}x23y
If [mn]x−1=[nm]x−5\Big[\dfrac{m}{n}\Big]^{x-1} = \Big[\dfrac{n}{m}\Big]^{x-5}[nm]x−1=[mn]x−5, the value of x is:
3
-3
13\dfrac{1}{3}31
−13-\dfrac{1}{3}−31
[17]−3×7−1×[149]\Big[\dfrac{1}{7}\Big]^{-3} \times 7^{-1} \times \Big[\dfrac{1}{49}\Big][71]−3×7−1×[491] is equal to:
-1
17\dfrac{1}{7}71
-7
1
