If 2x+y=8 and (x+y)=(3/2)log_(10) 2.log_8 36 then x^2 +3y=... a. 28 b. 22 c. 20 d. 16 e. 12


Question and Answers Forum

All Questions Topic List
Logarithms Questions
Previous in All Question Next in All Question
Previous in Logarithms Next in Logarithms
Question Number 66354 by gunawan last updated on 13/Aug/19

$If 2 x + y = 8 and$ $(x + y) = \frac{3}{2} \log_{10} 2 . \log_{8} 36$ $then x^{2} + 3 y = . . .$ $a . 28$ $b . 22$ $c . 20$ $d . 16$ $e . 12$
Commented by kaivan.ahmadi last updated on 13/Aug/19
$(3/2)log_(10) 2.log_8 36=(3/2)log_(10) 2.(2/3)log_2 6= log_(10) 2.log_2 6=((log_2 6)/(log_2 10))=log_(10) 6=x+y⇒ { ((2x+y=8)),((x+y=log_(10) 6)) :}⇒ { ((2x+y=8)),((−2x−2y=−2log_(10) 6)) :}⇒ y=−8+2log_(10) 6 and 2x−8+2log_(10) 6=8⇒x=8−log_(10) 6 so x^2 +3y=64−16log_(10) 6+(log_(10) 6)^2 −24+6log_(10) 6= (log_(10) 6)^2 −10log_(10) 6+40=$
$\frac{3}{2} {l o g}_{10} 2 . {l o g}_{8} 36 = \frac{3}{2} {l o g}_{10} 2 . \frac{2}{3} {l o g}_{2} 6 =$ ${l o g}_{10} 2 . {l o g}_{2} 6 = \frac{{l o g}_{2} 6}{{l o g}_{2} 10} = {l o g}_{10} 6 = x + y \Rightarrow$ ${\begin{cases} 2 x + y = 8 \\ x + y = {l o g}_{10} 6 \end{cases} \Rightarrow {\begin{cases} 2 x + y = 8 \\ - 2 x - 2 y = - 2 {l o g}_{10} 6 \end{cases} \Rightarrow$ $y = - 8 + 2 {l o g}_{10} 6$ $a n d$ $2 x - 8 + 2 {l o g}_{10} 6 = 8 \Rightarrow x = 8 - {l o g}_{10} 6$ $s o$ $x^{2} + 3 y = 64 - 16 {l o g}_{10} 6 + {({l o g}_{10} 6)}^{2} - 24 + 6 {l o g}_{10} 6 =$ ${({l o g}_{10} 6)}^{2} - 10 {l o g}_{10} 6 + 40 =$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum