Download Logarithm Questions for XAT\u00a0 PDF \u2013 XAT Logarithm questions PDF by Cracku. Practice XAT solved Logarithm Questions paper tests, which are the practice question to have a firm grasp on the Logarithm topic in the XAT exam. Top 20 very Important Logarithm Questions for XAT based on asked questions in previous exam papers.\u00a0 This PDF contains the most important Logarithm questions with detailed video solutions. Click on the below link to download the PDF of the Logarithm questions for XAT 2023.<\/p>\n
Download Logarithm Questions for XAT 2023<\/a><\/p>\n Enroll to XAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>If $log_3 2, log_3 (2^x – 5), log_3 (2^x – 7\/2)$ are in arithmetic progression, then the value of x is equal to<\/p>\n a)\u00a05<\/p>\n b)\u00a04<\/p>\n c)\u00a02<\/p>\n d)\u00a03<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $2 log (2^x – 5) = log 2 + log (2^x – 7\/2)$ Question 2:\u00a0<\/b>Let $u = ({\\log_2 x})^2 – 6 {\\log_2 x} + 12$ where x is a real number. Then the equation $x^u = 256$, has<\/p>\n a)\u00a0no solution for x<\/p>\n b)\u00a0exactly one solution for x<\/p>\n c)\u00a0exactly two distinct solutions for x<\/p>\n d)\u00a0exactly three distinct solutions for x<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $x^u = 256$<\/p>\n Taking log to the base 2 on both the sides,<\/p>\n $u * \\log_{2}{x} = \\log_{2}{256}$<\/p>\n =>$[({\\log_2 x})^2 – 6 {\\log_2 x} + 12] * \\log_{2}{x} = 8$<\/p>\n $(log_2 x)^3 – 6(log_2 x)^2 + 12log_2 x = 8$<\/p>\n Let $log_2 x = t$<\/p>\n $t^3 – 6t^2 +12t – 8 = 0$<\/p>\n $(t-2)^3 = 0$<\/p>\n Therefore, $log_2 x = 2$<\/p>\n => $x = 4$ is the only solution<\/p>\n Hence, option B is the correct answer.<\/p>\n Question 3:\u00a0<\/b>If $log_y x = (a*log_z y) = (b*log_x z) = ab$, then which of the following pairs of values for (a, b) is not possible?<\/p>\n a)\u00a0(-2, 1\/2)<\/p>\n b)\u00a0(1,1)<\/p>\n c)\u00a0(0.4, 2.5)<\/p>\n d)\u00a0($\\pi$, 1\/ $\\pi$)<\/p>\n e)\u00a0(2,2)<\/p>\n 3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $log_y x = ab$ Question 4:\u00a0<\/b>If x >= y and y > 1, then the value of the expression $log_x (x\/y) + log_y (y\/x)$ can never be<\/p>\n a)\u00a0-1<\/p>\n b)\u00a0-0.5<\/p>\n c)\u00a00<\/p>\n d)\u00a01<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $log_x (x\/y) + log_y (y\/x)$ = $1 – log_x (y) + 1 – log_y (x)$ Question 5:\u00a0<\/b>If $f(x) = \\log \\frac{(1+x)}{(1-x)}$, then f(x) + f(y) is<\/p>\n a)\u00a0$f(x+y)$<\/p>\n b)\u00a0$f{\\frac{(x+y)}{(1+xy)}}$<\/p>\n c)\u00a0$(x+y)f{\\frac{1}{(1+xy)}}$<\/p>\n d)\u00a0$\\frac{f(x)+f(y)}{(1+xy)}$<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n If $f(x) = \\log \\frac{(1+x)}{(1-x)}$ then $f(y) = \\log \\frac{(1+y)}{(1-y)}$<\/p>\n Also Log (A*B)= Log A + Log B<\/p>\n f(x)+f(y) = $ \\log \\frac{(1+x)(1+y)}{(1-x)(1-y)}$<\/p>\n =$\\log\\frac{\\left(1+xy\\ +x\\ +y\\right)}{\\left(1+xy-x-y\\right)}$<\/p>\n Dividing numberator and denominator by (1+xy)<\/p>\n $\\log\\frac{\\frac{\\left(1+xy\\ +x\\ +y\\right)}{1+xy}}{\\frac{\\left(1+xy-x-y\\right)}{1+xy}}$<\/p>\n =$\\log\\frac{\\frac{1+xy\\ }{1+xy}+\\frac{\\left(x+y\\right)}{1+xy}}{\\frac{1+xy\\ }{1+xy}-\\frac{\\left(x+y\\right)}{1+xy}}$<\/p>\n =\u00a0$\\log { \\frac{1+ \\frac{(x+y)}{(1+xy)}}{1- \\frac{(x+y)}{(1+xy)}}}$<\/p>\n Hence option B.<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>If $\\log_{2}{\\log_{7}{(x^2 – x+37)}}$ = 1, then what could be the value of \u2018x\u2019?<\/p>\n a)\u00a03<\/p>\n b)\u00a05<\/p>\n c)\u00a04<\/p>\n d)\u00a0None of these<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $\\log_{2}{\\log_{7}{(x^2 – x+37)}}$ = 1<\/p>\n $\\log_{7}{(x^2 – x+37)}$ = $2$<\/p>\n $(x^2 – x+37)$ = $7^{2}$<\/p>\n Given eq. can be reduced to $x^2 – x + 37 = 49$<\/p>\n So x can be either -3 or 4.<\/p>\n Question 7:\u00a0<\/b>If $\\log_{2}{x}.\\log_{\\frac{x}{64}}{2}=\\log_{\\frac{x}{16}}{2}$. Then x is<\/p>\n a)\u00a02<\/p>\n b)\u00a04<\/p>\n c)\u00a016<\/p>\n d)\u00a012<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\log_{2}{x}.\\log_{\\frac{x}{64}}{2}=\\log_{\\frac{x}{16}}{2}$<\/p>\n i.e. $\\frac{log{x}}{log{2}} * \\frac{log_{2}}{log{x}-log{64}} = \\frac{log{2}}{log{x}-log{16}}$<\/p>\n i.e.\u00a0$\\frac{log{x} * (log{x}-log{16})}{log{x}-log{64}}$ = $\\log{2}$<\/p>\n let t = log x<\/p>\n Therefore,\u00a0\u00a0$\\frac{t * (t-log{16})}{t-log{64}}$ = $\\log{2}$<\/p>\n $t^2-4*log 2*t = t*log 2-6*(log 2)^2$<\/p>\n I.e.\u00a0$t^2-5*log 2*t-6*(log 2)^2$ = 0<\/p>\n I.e.\u00a0$t^2-3*log 2*t-2*log 2*t-6*(log 2)^2$ = 0<\/p>\n i.e. $t*(t-3*log 2)-2*log 2*(t-3*log 2)$ = 0<\/p>\n i.e $t=2*log 2$ or $t=3*log 2$<\/p>\n i.e $log x=log 4$ or $log x=log 8$<\/p>\n therefore $x=4$ or $8$<\/p>\n therefore our answer is option ‘B’<\/p>\n Question 8:\u00a0<\/b>What is the value of $\\sqrt{\\frac{a}{b}}$, If $\\log_{4}\\log_{4}4^{a-b}=2\\log_{4}(\\sqrt{a}-\\sqrt{b})+1$<\/p>\n a)\u00a0-5\/3<\/p>\n b)\u00a02<\/p>\n c)\u00a05\/3<\/p>\n d)\u00a01<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sqrt{\\frac{a}{b}}$, If $\\log_{4}\\log_{4}4^{a-b}=2\\log_{4}(\\sqrt{a}-\\sqrt{b})+\\log_{4}{4}$<\/p>\n i.e. $\\log_{4}\\log_{4}4^{a-b}=\\log_{4}((\\sqrt{a}-\\sqrt{b})^2)*4$<\/p>\n i.e. $\\log_{4}4^{a-b}=((\\sqrt{a}-\\sqrt{b})^2)*4$<\/p>\n i.e. (a-b)*$\\log_{4}4=((\\sqrt{a}-\\sqrt{b})^2)*4$<\/p>\n i.e. a-b = 4a+4b-8$\\sqrt{ab}$<\/p>\n i.e. 3a + 5b – 8$\\sqrt{ab}$ = 0<\/p>\n i.e. $3\\sqrt\\frac{a}{b}^2$ – 8$\\sqrt\\frac{a}{b}$+5 = 0<\/p>\n put\u00a0$\\sqrt\\frac{a}{b}$ = t<\/p>\n therefore 3$t^2$ – 8t + 5 = 0<\/p>\n solving we get t = 1 or t = $\\frac{5}{3}$<\/p>\n i.e.\u00a0$\\sqrt\\frac{a}{b}$ = 1 or\u00a0$\\frac{5}{3}$<\/p>\n but if\u00a0$\\sqrt\\frac{a}{b}$ = 1 then a=b then $\\log_{4}(\\sqrt{a}-\\sqrt{b})$ will become indefinite<\/p>\n Therefore\u00a0\u00a0$\\sqrt\\frac{a}{b}$ =\u00a0$\\frac{5}{3}$<\/p>\n Therefore our answer is option ‘C’<\/p>\n Question 9:\u00a0<\/b>Find the value of x from the following equation: a)\u00a02\/7<\/p>\n b)\u00a07\/2<\/p>\n c)\u00a09\/2<\/p>\n d)\u00a0None of the above<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\log_{10}{3}+\\log_{10}(4x+1)=\\log_{10}(x+1)+1$ can be written as<\/p>\n $\\log_{10}{3}+\\log_{10}(4x+1)=\\log_{10}(x+1)+\\log_{10}{10}$<\/p>\n We know that\u00a0$\\log_{10}{a}+\\log_{10}{b}=\\log_{10}{ab}$<\/p>\n $\\log_{10}{3*(4x+1)}=\\log_{10}{(x+1)*10}$<\/p>\n $12x+3=10x+10$<\/p>\n $x=7\/2$. Hence, option B is the correct answer.<\/p>\n Question 10:\u00a0<\/b>If $\\log{3}, log(3^{x} – 2)$ and $log (3^{x}+ 4)$ are in arithmetic progression, then x is equal to<\/p>\n a)\u00a0$\\frac{8}{3}$<\/p>\n b)\u00a0$\\log_{3}{8}$<\/p>\n c)\u00a0$\\log_{2}{3}$<\/p>\n d)\u00a0$8$<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n If $log{3}, log(3^{x} – 2)$ and $log (3^{x}+ 4)$ are in arithmetic progression Question 11:\u00a0<\/b>If $log_{10} x – log_{10} \\sqrt[3]{x} = 6log_{x}10$ then the value of x is<\/p>\n a)\u00a010<\/p>\n b)\u00a030<\/p>\n c)\u00a0100<\/p>\n d)\u00a01000<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\log_{10} x – \\log_{10} \\sqrt[3]{x} = 6\\log_{x}10$ Question 12:\u00a0<\/b>The value of $\\text{log}_{7} \\text{log}_{7} \\sqrt{7(\\sqrt{7\\sqrt{7}})}$<\/p>\n a)\u00a07<\/p>\n b)\u00a0$\\text{log}_7$ 2<\/p>\n c)\u00a0$1-3 \\text{log}_2$ 7<\/p>\n d)\u00a0$1-3 \\text{log}_7$ 2<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\text{log}_{7} \\text{log}_{7} \\sqrt{7(\\sqrt{7\\sqrt{7}})}$ Question 13:\u00a0<\/b>$log_{13} log_{21} (\\sqrt{x+21}+ \\sqrt{x} ) =0 $ then the value of x is<\/p>\n a)\u00a021<\/p>\n b)\u00a013<\/p>\n c)\u00a081<\/p>\n d)\u00a0None of the above<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $log_{13} log_{21} (\\sqrt{x+21}+ \\sqrt{x} ) =0 $ Question 14:\u00a0<\/b>The value of $\\text{log}_{7} \\text{log}_{7} \\sqrt{7(\\sqrt{7\\sqrt{7}})}$<\/p>\n a)\u00a07<\/p>\n b)\u00a0$\\text{log}_7$ 2<\/p>\n c)\u00a0$1-3 \\text{log}_2$ 7<\/p>\n d)\u00a0$1-3 \\text{log}_7$ 2<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\text{log}_{7} \\text{log}_{7} \\sqrt{7(\\sqrt{7\\sqrt{7}})}$ Question 15:\u00a0<\/b>If $log_{25}{5}$ = a and $log_{25}{15} $ = b, then the value of $log_{25}{27}$ is:<\/p>\n a)\u00a03(b+a)<\/p>\n b)\u00a03(1-b-a)<\/p>\n c)\u00a03(a+b-1)<\/p>\n d)\u00a03(1-b+a)<\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $log_{25}{5}$ = a<\/p>\n => a=1\/2<\/p>\n $log_{25}{15}$ = $log_{25}{3}+log_{25}{5}$ = b<\/p>\n $\\frac{1}{2} \\log_{5}{3} + \\frac{1}{2}$ = b<\/p>\n $\\log_{5}{3}$ = 2(b –\u00a0 $\\frac{1}{2})$………….(i)<\/p>\n $log_{25}{27}$ = $\\frac{3}{2} \\log_{5}{3}$………(ii)<\/p>\n Replacing $\\log_{5}{3}$ = 2(b –\u00a0 $\\frac{1}{2})$ in (ii) we get<\/p>\n $log_{25}{27}$ = 3(b –\u00a0 $\\frac{1}{2}$)<\/p>\n We can write -$\\frac{1}{2}$ as (- 1 + $\\frac{1}{2}$) or (-1 + a)<\/p>\n So,\u00a0$log_{25}{27}$ = 3(b + a – 1)<\/p>\n Hence, option C is the correct answer.<\/p>\n Question 16:\u00a0<\/b>Find the value of x which satisfies the following equation $4log_7 (x – 8) = log_3 81$<\/p>\n a)\u00a08<\/p>\n b)\u00a018<\/p>\n c)\u00a020<\/p>\n d)\u00a0None of the above<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $4log_7 (x – 8) = log_3 81$<\/p>\n $4log_7 (x – 8) = log_3 3^{4}$<\/p>\n $4log_7 (x – 8) =4*log_3 3$<\/p>\n $log_7 (x – 8) =1$<\/p>\n $(x – 8) =7^{1}$<\/p>\n $x = 7+8$<\/p>\n $x = 15$<\/p>\n Question 17:\u00a0<\/b>$(1+5)\\log_{e}3+\\frac{(1+5^{2})}{2!}(\\log_{e}3)^{2}+\\frac{(1+5^{3})}{3!}(\\log_{e}3)^{3}+…$<\/p>\n a)\u00a012<\/p>\n b)\u00a0244<\/p>\n c)\u00a0243<\/p>\n d)\u00a0245<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Splitting the above mentioned series into two series<\/p>\n A = $\\log_{e}3+\\frac{1}{2!}(\\log_{e}3)^{2}+\\frac{1}{3!}(\\log_{e}3)^{3}+…$<\/p>\n B = $5\\log_{e}3+\\frac{5^{2}}{2!}(\\log_{e}3)^{2}+\\frac{5^{3}}{3!}(\\log_{e}3)^{3}+…$<\/p>\n We know that $e^{x}$ =$1+x+\\frac{x^{2}}{2!}+\\frac{x^{3}}{3!}+…$<\/p>\n So\u00a0\u00a0$e^{x}-1$ = $x+\\frac{x^{2}}{2!}+\\frac{x^{3}}{3!}+…$<\/p>\n On solving two series A and B<\/p>\n A = $\\log_{e}3+\\frac{1}{2!}(\\log_{e}3)^{2}+\\frac{1}{3!}(\\log_{e}3)^{3}+…$ =$e^{\\log_{e}3}-1$ = $3-1$ =$2$<\/p>\n B = $5\\log_{e}3+\\frac{5^{2}}{2!}(\\log_{e}3)^{2}+\\frac{5^{3}}{3!}(\\log_{e}3)^{3}+…$<\/span>=$e^{\\log_{e}3^{5}}-1$=$3^{5}-1$=$242$<\/p>\n A+B = $2 + 242$ = $244$<\/p>\n Question 18:\u00a0<\/b>Suppose, $\\log_3 x = \\log_{12} y = a$, where $x, y$ are positive numbers. If $G$ is the geometric mean of x and y, and $\\log_6 G$ is equal to<\/p>\n a)\u00a0$\\sqrt{a}$<\/p>\n b)\u00a02a<\/p>\n c)\u00a0a\/2<\/p>\n d)\u00a0a<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We know that $\\log_3 x = a$ and $\\log_{12} y=a$ Hence, $G=6^a$ Or, $\\log_6 G = a$<\/p>\n Question 19:\u00a0<\/b>The value of $\\log_{0.008}\\sqrt{5}+\\log_{\\sqrt{3}}81-7$ is equal to<\/p>\n a)\u00a01\/3<\/p>\n b)\u00a02\/3<\/p>\n c)\u00a05\/6<\/p>\n d)\u00a07\/6<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $\\log_{0.008}\\sqrt{5}+\\log_{\\sqrt{3}}81-7$<\/p>\n $81 = 3^4$ and $0.008 = \\frac{8}{1000} = \\frac{2^{3}}{10^{3}} = \\frac{1}{5^{3}} = 5^{-3} $<\/p>\n Hence,<\/p>\n $\\log_{0.008}\\sqrt{5}+ 8 -7 $<\/p>\n $ \\log_{5^{-3}}5^{\\frac{1}{2}}+ 8 -7 $<\/p>\n $\\frac{log 5^{0.5}}{log 5^{-3}} + 1$<\/p>\n $ – \\frac{1}{6} + 1$<\/p>\n = $\\frac{5}{6}$<\/p>\n Question 20:\u00a0<\/b>If x is a real number such that $\\log_{3}5= \\log_{5}(2 + x)$, then which of the following is true?<\/p>\n a)\u00a00 < x < 3<\/p>\n b)\u00a023 < x < 30<\/p>\n c)\u00a0x > 30<\/p>\n d)\u00a03 < x < 23<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $1 < \\log_{3}5 < 2$
\nLet $2^x = t$
\n=> $(t-5)^2 = 2(t-7\/2)$
\n=> $t^2 + 25 – 10t = 2t – 7$
\n=> $t^2 – 12t + 32 = 0$
\n=> t = 8, 4
\nTherefore, x = 2 or 3, but $2^x$ > 5, so x = 3<\/p>\n
\n$a*log_z y = ab$ => $log_z y = b$
\n$b*log_x z = ab$ => $log_x z = a$
\n$log_y x$ = $log_z y * log_x z$ => $log x\/log y$ = $log y\/log z * log z\/log x$
\n=> $\\frac{log x}{log y} = \\frac{log y}{log x}$
\n=> $(log x)^2 = (log y)^2$
\n=> $log x = log y$ or $log x = -log y$
\nSo, x = y or x = 1\/y
\nSo, ab = 1 or -1
\nOption 5) is not possible<\/p>\n
\n= $2 – (log_x y + 1\/log_x y)$ <= 0 (Since $log_x y + 1\/log_x y$ >= 2)
\nSo, the value of the expression cannot be 1.<\/p>\n
\n$\\log_{10}{3}+\\log_{10}(4x+1)=\\log_{10}(x+1)+1$<\/p>\n
\nThen, $2*log(3^{x} – 2) = log{3}+log (3^{x}+ 4)$
\nThus, $log{(3^{x} – 2)^2} = log{3(3^x+4)}$
\nThus, $(3^{x} – 2)^2 = 3(3^x+4)$
\n=> $3^{2x} – 4*3^x +4 = 3*3^x + 12$
\n=> $3^{2x} – 7*3^x – 8 = 0$
\n=> $(3^x+1)*(3^x-8) = 0$
\nBut $3^x+1 \\neq 0$
\nThus, $3^x = 8$
\nHence, $x = log_{3}{8}$
\nHence, option B is the correct answer.<\/p>\n
\nThus, $\\dfrac{\\log {x}}{\\log {10}}$ – $\\dfrac{1}{3}*\\dfrac{\\log {x}}{\\log {10}}$ = $6*\\dfrac{\\log{10}}{\\log{x}}$
\n=> $\\dfrac{2}{3}*\\dfrac{\\log {x}}{\\log {10}}$ = $6*\\dfrac{\\log{10}}{\\log{x}}$
\nThus, => $\\dfrac{1}{9}*(\\log{x})^2 = (\\log{10})^2=1$
\nThus,\u00a0$(\\log{x})^2 = 9$
\nThus $\\log x = 3$ or $-3$
\nThus, $ x = 1000$ or $\\dfrac{1}{1000}$
\nFrom amongst the given options, 1000 is the correct answer.
\nHence, option D is the correct answer.<\/p>\n
\n=\u00a0$\\text{log}_{7} [\\frac{1}{2} (\\text{log}_{7} 7+\\text{log}_{7} \\sqrt{7(\\sqrt{7})})]$
\n= $\\text{log}_{7} [\\frac{1}{2} (1 +\\frac{1}{2}\\text{log}_{7}{7(\\sqrt{7})})]$
\n= $\\text{log}_{7} [\\frac{1}{2} (1 +\\frac{1}{2}(1+1\/2))]$
\n= $\\text{log}_{7}\\frac{7}{8}$
\n= $\\text{log}_{7}7 – \\text{log}_{7}8$
\n= $1-3 \\text{log}_{7}2$
\nHence, option D is the correct answer.<\/p>\n
\nThus, $log_{21} (\\sqrt{x+21}+ \\sqrt{x} ) = 1 $
\nThus, $(\\sqrt{x+21}+ \\sqrt{x} ) = 21 $
\nLet, $\\sqrt{x} = t$
\nThus, $x = t^2$
\nThus, $x+21 = t^2+21$
\nThus, $\\sqrt{t^2+21}+t = 21$
\nThus, $(t^2+21) = (21-t)^2$
\n=> $t^2 + 21 = 441 – 42t + t^2$
\n=> $42t = 420$
\nHence, $t = 10$
\nHence, option D is the correct answer.<\/p>\n
\n=\u00a0$\\text{log}_{7} [\\frac{1}{2} (\\text{log}_{7} 7+\\text{log}_{7} \\sqrt{7(\\sqrt{7})})]$
\n= $\\text{log}_{7} [\\frac{1}{2} (1 +\\frac{1}{2}\\text{log}_{7}{7(\\sqrt{7})})]$
\n= $\\text{log}_{7} [\\frac{1}{2} (1 +\\frac{1}{2}(1+1\/2))]$
\n= $\\text{log}_{7}\\frac{7}{8}$
\n= $\\text{log}_{7}7 – \\text{log}_{7}8$
\n= $1-3 \\text{log}_{7}2$
\nHence, option D is the correct answer.<\/p>\n
\nHence, $x = 3^a$ and $y=12^a$
\nTherefore, the geometric mean of $x$ and $y$ equals $\\sqrt{x \\times y}$
\nThis equals $\\sqrt{3^a \\times 12^a} = 6^a$<\/p>\n
\n=> $ 1 < \\log_{5}(2 + x) < 2 $
\n=> $ 5 < 2 + x < 25$
\n=> $ 3 < x < 23$<\/p>\n