A.logπ₯ B.1 C.0 D.12logπ₯ E.2
- Solve the inequality xlog100.1>log1010.
A. x<-1 B. x<1 C. x>1 D. x>100 E. x>-1 - If logπ+π= logπβlogπ , then p=β¦β¦β¦.
A. p=q=1 B. p=π1βπ C. p=π21βπ D. p= π1+π E. p=π21+π
25.If logπ=5, logπ =3, then the value of ππ is
A. 53 B. 2 C. 8 D. log53 E.100 - Given that logπ2 =0.301 and logπ3 =0.477, then =β¦β¦..
A. 0.125 B. -0.125 C. 0.301 D. -1.125 E. 1.125 - If 2logπ8βlogπ4=2, then p=β¦β¦β¦.
A. 4 B. -4 C. 4 (or) 2 D. 4 (or) -4 E. 2 - log19π₯β1π₯+2 = 12; x=β¦β¦β¦
A. 12 B. 32 C. 52 D. 72 E. 92 - log39π₯β22= x+2 ; x=β¦β¦β¦β¦
A. log113 B. log311 C.log3 D.log11 E. 0
30.If log2=m, log3=n, then log720 =β¦β¦β¦.
A. m+n+1 B. 3m+n+1 C. 2m+3n+1 D. 3m+2n+1 E. 3m+2n-1 - log29 =a , log26=β¦β¦β¦..
A. 1π+2 B.π+22 C. -a D. a+12 E. 2a - log0.040.4=β¦β¦.
A. -3 B. -2 C. -1 D. 1 E.4 - log55+log31+log416=β¦β¦.
A. 0 B. 1 C. 2 D. 3 E.4 - If log2.7 =0.431 , then log2.7 =β¦β¦β¦
A.1 .431 B.-0.215 C.0.2155 D. 0.8