The logarithm is an important topic in the Quant section of the SNAP Exam. Quant is a scoring section in SNAP, so it is advised to practice as much as questions from quant. This article provides some of the most important Logarithm Questions for SNAP. One can also download this Free Logarithm Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Logarithm questions in the SNAP exam. Utilize this PDF practice set, <\/strong>which is one of the best sources for practising.<\/p>\n Download Logarithm Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 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\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/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>$\\log_{5}{2}$ is<\/p>\n a)\u00a0An integer<\/p>\n b)\u00a0A rational number<\/p>\n c)\u00a0A prime number<\/p>\n d)\u00a0An irrational number<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let $\\log_{5}{2}$ = y<\/p>\n Let us assume\u00a0\u00a0$\\log_{5}{2}$ is a rational number.<\/p>\n $\\log_{5}{2}$ = p\/q, where p and q are co primes.<\/p>\n 5^(p\/q)=2 => 5^p=2^q.<\/p>\n 5^p=5*5*5*5*5*5*5………………p times<\/p>\n 2^p=2*2*2*2*2*2*2………………q times<\/p>\n No value of p and q can satisfy the equation. Hence y is an irrational number.<\/p>\n Question 10:\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 10)\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
\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