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Question Number 121783 by Ar Brandon last updated on 11/Nov/20

The number of integer values of x satisfying the inequality 2x+1<2log_2 (x+3) is ___.

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integer}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{satisfying}\: \\ $$ $$\mathrm{the}\:\mathrm{inequality}\:\mathrm{2x}+\mathrm{1}<\mathrm{2log}_{\mathrm{2}} \left(\mathrm{x}+\mathrm{3}\right)\:\mathrm{is}\:\_\_\_. \\ $$

Answered by TANMAY PANACEA last updated on 11/Nov/20

x=1 LHS=3 RHS=4 RHS>LHS x=−1 LHS =−1 RHS=2 RHS>LHS x=5 LHS=11 RHS=6 LHS>RHS x=13 L=27 R=8 L>R x=29 L=59 R=10 L>R so x=1 and −1 satisfy the eqn

$$ \\ $$ $${x}=\mathrm{1}\:\:{LHS}=\mathrm{3}\:\:{RHS}=\mathrm{4}\:\:{RHS}>{LHS} \\ $$ $${x}=−\mathrm{1}\:{LHS}\:=−\mathrm{1}\:\:{RHS}=\mathrm{2}\:\:{RHS}>{LHS} \\ $$ $${x}=\mathrm{5}\:{LHS}=\mathrm{11}\:\:{RHS}=\mathrm{6}\:\:{LHS}>{RHS} \\ $$ $${x}=\mathrm{13}\:\:{L}=\mathrm{27}\:\:{R}=\mathrm{8}\:\:{L}>{R} \\ $$ $${x}=\mathrm{29}\:\:{L}=\mathrm{59}\:\:{R}=\mathrm{10}\:\:{L}>{R} \\ $$ $$\boldsymbol{{so}}\:\:\:\:\boldsymbol{{x}}=\mathrm{1}\:\boldsymbol{{and}}\:−\mathrm{1}\:\boldsymbol{{satisfy}}\:\boldsymbol{{the}}\:\boldsymbol{{eqn}} \\ $$

Commented byAr Brandon last updated on 11/Nov/20

Thanks Sir. But there are 4 integers according to the answer guide.

Commented byDwaipayan Shikari last updated on 11/Nov/20

0 and −2 (Another two)

$$\mathrm{0}\:{and}\:−\mathrm{2}\:\left({Another}\:{two}\right) \\ $$

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