Question

# Find the values of x in each of the following: (i)25x÷2x=5√220 (ii)(23)4=(22)x (iii)(35)x(53)2x=12527 (iv)5x−2×32x−3=135 (v)2x−7×5x−4=1250 (vi)(3√4)2x+12=132 (vii)52x+3=1 (viii)   (13)√x=44−34−6 (ix)(√35)x+1=12527

Solution

## (i)25x÷2x=5√220⇒25x2x=(220)15⇒25x−x=2205⇒24x=2205Comparing,we get4x=205⇒x=205×4=1∴x=1 (ii)(23)4=(22)x⇒23×4=22×x⇒212=22xComparing,we get2x=12⇒x=122=6∴x=6 (iii)(35)x(53)2x=12527⇒(53)−x(53)2x=5333⇒(53)−x+2x=(53)3Comparing,we get−x+2x=3⇒x=3∴x=3 (iv)5x−2×32x−3=1355x−2×32x−3=1355x.5−2.32x.3−3=1355x.32x52×33=1355x.32x=135×52×335x×3x×3x=135×25×27(5×3×3)x=3×3×3×5×5×5×3×3×3(45)x=(3×5×3)3=(45)3Comparing,we getx=3 (v)2x−7×5x−4=12502x−7×5x−4=1250⇒2x.2−7.5x.5−4=2×5×5×5×5⇒2x.5x27×54=2×5×5×5×5⇒(10)x=2×5×5×5×5×2×2×2×2×2×2×2×5×5×5×5⇒(10)x=28×58=(10)8Comparing,we getx=8 (vi)(3√4)2x+12=132⇒[(22)13]2x+12=125⇒223(2x+12)=2−5⇒23(2x+12)=−5⇒43x+13=−5⇒8x+2=−5×6⇒8x=−30−2⇒x=−328=−4 (vii)52x+3=1=50(∵a0=1)∴2x+3=0⇒2x=−3∴x=−32 (viii)   (13)√x=44−34−6=256−81−6=169=(13)2∴√x=2Squaring,x=4 (ix)(√35)x+1=12527⇒(35)x+12=(35)−3          {∵am=1a−m}x+12=−3⇒x+1=−6⇒x=−7∴x=−7 MathematicsRD SharmaStandard IX

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