6^log36(√3－2)^2+6^log36(√3+2)^2=?(lg5)^2+lg2×lg50=?2^log65×3^log65=?lg5-1=?

Question

6^log36 (√3－2)^2+6^log36 (√3+2)^2=?
(lg5)^2+lg2×lg50=?
2^log6 5×3^log6 5=?
lg5-1=?

Answer 1

6^log36 (√3－2)^2+6^log36 (√3+2)^2
=6^{[log6 (√3－2)^2]/[log6 (36)]}+6^{[log6 (√3+2)^2]/[log6 (36)]}
=6^{[log6 (√3－2)^2]/[log6 (6^2)]}+6^{[log6 (√3+2)^2]/[log6 (6^2)]}
=6^{[log6 (√3－2)^2]/[2log6 (6)]}+6^{[log6 (√3+2)^2]/[2log6 (6)]}
=6^{[log6 (√3－2)^2]/2}+6^{[log6 (√3+2)^2]/2}
=6^{[2log6 (√3－2)]/2}+6^{[2log6 (√3+2)]/2}
=6^[log6 (√3－2)]+6^[log6 (√3+2)]
=√3－2+√3+2
=2√3
(lg5)^2+lg2×lg50
=(lg5)^2+lg2×(lg5×10)
=(lg5)^2+lg2×(lg5+lg10)
=(lg5)^2+lg2×(lg5+1)
=(lg5)^2+lg2×lg5+lg2
=(lg5)(lg5+lg2)+lg2
=(lg5)(lg5×2)+lg2
=(lg5)(lg10)+lg2
=(lg5)(1)+lg2
=lg5+lg2
=lg10
=1;
2^log6 5×3^log6 5
=(2×3)^log6 5
=6^log6 5
=5
lg5-1
=lg5-lg10
=-(lg10-lg5)
=-lg(10/5)
=-lg2