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Given Solution $s0$ $1.$ $7^{x}=\dfrac {1} {49}$ $2$ $5^{3x+8}=25^{2x}$ $3.4^{x+1}=\dfrac {1} {64}$ $4.$ $10^{x}>100^{-2x-5}$
10th-13th grade
Calculus
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Qanda teacher - kamalsahoo
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thank you sir