11. Logarithm Solutions Part 1#
\(\log_{\sqrt{8}}x = \frac{10}{3} \Rightarrow \log_2^{\frac{3}{2}} x = \frac{10}{3}\)
\(\Rightarrow \frac{2}{3}\log_2 x = \frac{10}{3} \Rightarrow \log_2 x = 5\)
\(\Rightarrow x = 25\)
L.H.S. \(= \log_b a.\log_c b.\log_a c = \frac{\log a}{\log b}\frac{\log b}{\log c}\frac{\log c}{\log a} = 1\)
L.H.S. \(= \log_3 \log_2 \log_{\sqrt{5}} (625) = \log_3 \log_2 \log_{\sqrt{5}} 5^4\)
\(= \log_3 \log_2 8 = \log_3 \log_2 2^3 = \log_3 3 = 1\)
Given, \(a^2 + b^2 = 23ab \Rightarrow (a + b)^2 = 25ab\)
\(\Rightarrow \left(\frac{a + b}{5}\right) = \sqrt{ab}\)
Taking log on both sides, we get
\(\log \frac{a + b}{5} = \frac{1}{2}(\log a + \log b)\)
L.H.S. \(= 7\log \frac{16}{15} + 5\log \frac{25}{24} + 3\log \frac{81}{80}\)
\(= 7[\log 2^4 - \log(3.5)] + 5[\log 5^2 - \log 2.2.2.3] + 3[\log 3^4 - \log 2^4.5]\)
\(= 7[4\log2 - \log 3 - \log 5] + 5[2\log 5 - 3\log 2 -\log 3] + 3[4\log 3 - 4\log 2 -\log 5]\)
\(= \log 2 =\) R.H.S.
L.H.S. \(=\log\tan 1^{\circ} + \log\tan 2^{\circ} + \ldots + \log \tan 89^{\circ}\)
\(= \log(\tan 1^{\circ}.\tan 89^{\circ}).\log(\tan 2^{\circ}.\tan 88^{\circ}) \ldots \log \tan 45^{\circ}\)
\(= \log(\tan 1^{\circ}.\cot 1^{\circ}).\log(\tan 2^{\circ}.\cot 2^{\circ}) \ldots \log \tan 45^{\circ}\)
\(= \log 1.\log 1\ldots \log 1 = 0\)
\(\log_9 \tan \frac{\pi}{6} = \log_9 \frac{1}{\sqrt{3}} = \log_3^2 3^{-\frac{1}{2}}\)
\(= \frac{1}{2}\log_3 3^{-\frac{1}{2}} = \frac{1}{2}.-\frac{1}{2} = -\frac{1}{4}\)
\(\frac{\log_{a^2}b}{\log_{\sqrt{a}}(b)^2} = \frac{\frac{1}{2}\log_a b}{2.2\log_a b}\)
\(= \frac{\log_a b}{8\log_a b} = \frac{1}{8}\)
\(\log_{\sqrt{5}}.008 = 2\log_5 \frac{8}{1000} = 2[\log_5 8 - \log_5 1000]\)
\(= 2[\log_5 8 - \log_5 8 - \log 125] = -6\)
\(\log_{2\sqrt{3}}144 = \log_{2\sqrt{3}}12^2 = \log_{2\sqrt{3}}(2\sqrt{3})^4 = 4\)
L.H.S. \(=\log_3 \log_2 \log_{\sqrt{3}} 81 = \log_3 \log_2 \log_{\sqrt{3}}(\sqrt{3})^8\)
\(= \log_3 \log_2 8 = \log_3 \log_2 2^3 = \log_3 3 = 1\)
L.H.S. \(= \log_a x \log_b y = \frac{\log x}{\log a}\frac{\log y}{\log b}\)
\(= \frac{\log x}{\log b}\frac{\log y}{\log a} = \log_b x \log_a y =\) R.H.S.
L.H.S. \(= \log_2 \log_2 \log_2 16 = \log_2 \log_2 \log_2 2^4 = \log_2 \log_2 4 = \log_2 \log_2 2^2 = \log_2 2 = 1 =\) R.H.S.
R.H.S. \(= \log_b x\log_c b \ldots \log_n m \log_a n = \frac{\log x}{\log b}\frac{\log b}{\log c}\ldots \frac{\log m}{\log n}\frac{\log n}{\log a}\)
\(= \frac{\log x}{\log a} = \log_a x =\) L.H.S.
R.H.S. \(= 10^x\log_{10}a = z\) (say)
Taking log of both sides with base \(10\)
\(\log_{10} z =x\log_{10}a = \log_{10}a^x \Rightarrow z = a^x =\) L.H.S.
Given, \(a^2 + b^2 = 7ab \Rightarrow a^2 + b^2 + 2ab = 9ab\)
\(\Rightarrow \left(\frac{a + b}{3}\right)^2 = ab\)
Taking log of both sides
\(\Rightarrow 2\log \left(\frac{a + b}{3}\right) = \log ab = \log a + \log b\)
\(\Rightarrow \log \left\{\frac{1}{3}(a + b)\right\} = \frac{1}{2}(\log a + \log b)\)
L.H.S. \(= \frac{\log a(\log_b a)}{\log b(\log_a b)} = \frac{\log a \frac{\log a}{\log b}}{\log b \frac{\log a}{\log b}}\)
\(= \frac{\log a\log a - \log a \log b}{\log b \log b - \log b \log a}\)
\(\frac{\log a(\log a - \log b)}{\log b(\log b - \log a)} = - \log_b a =\) R.H.S.
L.H.S. \(= \log(1 + 2 + 3) = \log 6 = \log 2.3 = 0 + \log 2 + \log 3\)
\(= \log 1 + \log 2 + \log 3[\because \log 1 = 0]\)
L.H.S. \(= 2\log(1 + 2 + 4 + 7 + 14) = 2\log 28 = \log 28^2 = \log 784\)
\(= \log 1*2*4*7*14 = \log 1 + \log 2 + \log 4 + \log 7 + \log 14\)
L.H.S. \(= \log 2 + 16\log \frac{16}{15} + 12\log \frac{25}{24} + 7\log \frac{81}{80} = 1\)
\(= \log 2 + 16[\log 16 - \log 15] + 12[\log 25 - \log 24] + 7[\log 81 - \log 80]\)
\(= \log 2 + 16[\log 2^4 - \log 3*5] + 12[\log 5^2 - \log 2^3*3] + 7[\log 3^4 - \log 2^4*5]\)
\(= \log 2 + 16[4\log 2 - \log 3 -\log 5] + 12[2\log 5 - 3\log 2 -\log 3] + 7[4\log 3 - 4\log 2 -\log 5]\)
\(= \log 2[1 + 64 - 36 - 28] + \log 3[28 - 16 - 12] + \log 5[24 - 7 - 16]\)
\(= \log 2 + \log 5 = \log 10 = 1\) [default base is \(10\) for common logarithms]
Given, \(\frac{\log_9 11}{\log_5 13} \div \frac{\log_3 11}{\log_{\sqrt{5}} 13}\)
\(= \frac{\log_{3^2} 11}{\log_5 13}\frac{\log_{5^{\frac{1}{2}}} 13}{\log_3 11}\)
\(= \frac{\frac{1}{2}\log_3 11}{\log_5 13}\frac{2\log 13}{\log_3 11} = 1\)
Given, \(3^{\sqrt{\log_3 2}} - 2^{\sqrt{\log_2 3}}\)
Taking log with base \(10,\) we get
\(=\sqrt{\log_3 2}\log 3 - \sqrt{\log_2 3}\log 2\)
\(= \sqrt{\frac{\log 2}{\log 3}(\log 3)^2} - \sqrt{\frac{\log 3}{\log 2}(\log 2)^2}\)
\(= \sqrt{\log 2\log 3} - \sqrt{\log 3\log 2} = 0\)
Given \(\log_{10} 343 = 2.5353 \Rightarrow \log_{10} 7^3 = 2.5353 \Rightarrow \log_{10} 7 = .8451\)
For, \(7^n > 10^5 \Rightarrow n\log_{10}7 = 5\Rightarrow n = \frac{5}{.8451}\)
Thus, least such integer is \(6\)
Since \(a, b, c\) are in G.P., we can write \(b^2 = ac\)
Taking log of both sides
\(2\log b = \log a + \log c\)
Thus, \(\log a, \log b, \log c\) are in A.P.
i.e. \(\frac{1}{\log a}, \frac{1}{\log b}, \frac{1}{\log c}\) are in H.P.
Multiplying each term with \(\log x\)
i.e. \(\frac{\log x}{\log a}, \frac{\log x}{\log b}, \frac{\log x}{\log c}\) are in H.P.
i.e. \(\log_a x, \log_b x, \log_c x\) are in H.P.
L.H.S. \(= \log \sin 8x = \log 2\sin 4x \cos 4x\)
\(= \log 2 + \log \sin 4x + \log \cos 4x\)
\(= \log 2 + \log 2\sin 2x\cos 2x + \log \cos 4x\)
\(= 2\log 2 + \log \sin 2x + \log \cos 2x + \log \cos 4x\)
\(= 2\log 2 + \log 2\sin x\cos x + \log \cos 2x + \log \cos 4x\)
\(= 3\log 2 + \log \sin x + \log \cos x + \log \cos 2x + \log \cos 4x\)
We have to prove \(xyz + 1 = 2yz\)
Dividing both sides by \(yz,\) we get
\(x + \frac{1}{yz} = 2\)
L.H.S. \(= \log_{2a}a + \frac{1}{\log_{3a}2a\log_{4a}3a}\)
\(= \frac{\log a}{\log 2a} + \frac{\log 3a}{\log 2a}\frac{\log 4a}{\log 3a}\)
\(=\frac{\log a}{\log 2a} + \frac{\log 4a}{\log 2a}\)
\(= \frac{\log a + \log 4a}{\log 2a} = \frac{\log (2a)^2}{\log 2a} = 2\)
We have to prove that \(\log_{c + b} a + \log_{c - b} a = 2\log_{c + b}a\log_{c -b}a\)
Dividing both sides by \(\log_{c + b}a\log_{c -b}a,\) we get
\(\frac{1}{\log_{c - b}a} + \frac{1}{\log_{c + b}a} = 2\)
\(\log_a(c - b) + \log_a(c + b) = 2\)
\(\log_a(c^2 - b^2) = 2\)
\(c^2 - b^2 = a^2\)
\(c^2 = a^2 + b^2\) which is true because \(c\) is hypotenuse and \(a\) and \(b\) are sided of a right angle triangle.
We have to prove that, \(x^xy^yz^z = 1\)
Taking log of both sides
\(x\log x + y\log y + z\log z = 0\)
Let \(\frac{\log x}{y - z} = \frac{\log y}{z - x} = \frac{\log z}{x - y} = k\)
\(\log x = k(y - z), \log y = k(z - x), \log z = k(x - y)\)
\(x\log x + y\log y + z\log z = k(xy - zx + yz - xy + zx - yz) = 0\)
Given, \(\frac{yz\log(yz)}{y + z} = \frac{zx\log(zx)}{z + x} = \frac{xy\log(xy)}{x + y}\)
Dividing each term by \(xyz,\) we get
\(\frac{\log(yz)}{x(y + z)} = \frac{\log(zx)}{y(z + x)} = \frac{\log(xy)}{z(x + y)}\)
\(\frac{\log y + \log z}{xy + zx} = \frac{\log z + \log x}{yz + xy} = \frac{\log x + \log y}{zx + yz} = k\) (let)
\(\log y + \log z = k(xy + zx), \log z + \log x = k(yz + zx), \log x + \log y = k(zx + yz)\)
Adding all these, we get
\(2(\log x + \log y + \log z) = 2k(xy + yz + xz)\)
\((\log x + \log y + \log z) = k(xy + yz + xz)\)
\(\therefore \log x = kyz \Rightarrow x\log x = kxyz\)
Similarly \(y\log y = kxyz\) and \(z\log z = kxyz\)
\(\therefore x\log x = y\log y = z\log z = kxyz\)
\(\Rightarrow x^x = y^y = z^z\)
We have to prove that \((yz)^{\log y - \log z}(zx)^{\log z - \log x}(xy)^{\log x -\log y} = 1\)
Taking log of both sides
\((\log y - \log z)(\log y + \log z) + (\log z - \log x)(\log z - \log x) + (\log x - \log y)(\log x + \log y) = 0\)
\((\log y)^2 - (\log z)^2 + (\log z)^2 - (\log x)^2 + (\log x)^2 - (\log y)^2 = 0\)
\(0 = 0\)
We have to prove that \(\frac{1}{\log_2 N} + \frac{1}{\log_3 N} + \ldots + \frac{1}{\log_{1988} N} = \frac{1}{\log_{1988!} N}\)
L.H.S. \(= \log_N 2 + \log_N 3 + \ldots + \log_N 1988\)
\(= \log_N (2.3.4\ldots 1988)\)
\(= \log_N 1988! = \frac{1}{\log_{1988!}N}\)
L.H.S. \(= \log(1 + x) + \log(1 + x^2) + \log(1 + x^4) \ldots\) to \(\infty\)
\(\log (1 + x + x^3 + x^4 + \ldots)\) to \(\infty\)
\(= \log \frac{1}{1 - x}~[\because |x| < 1]\)
\(= -\log(1 - x)\)
Let \(S_n = \frac{1}{\log_2 a} + \frac{1}{\log_4 a} + \ldots\) upto \(n\) terms`
\(S_n = \log_a 2 + \log_a 4 + \log_a 8 + \ldots\) upto \(n\) terms
\(S_n = (1 + 2 + 3 + \ldots + n)\log_a 2\)
\(= \frac{n(n + 1)}{2}\log_a 2\)
L.H.S. \(= \frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1}\)
\(= \frac{1}{\log_4 10 + \log_{4} 4} + \frac{1}{\log_{2}20 + \log_2 2} + \frac{1}{\log_5 8 + \log_5 5}\)
\(= \frac{1}{\log_4 40} + \frac{1}{\log_2 40} + \frac{1}{\log_4 40}\)
\(= \log_{40} 4 + \log_{40} 2 + \log_{40} 5 = \log_{40} 40 = 1\)
We have to prove that \(\frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} = 1\)
L.H.S. \(= \frac{1}{=log_a{bc} + 1} + \frac{1}{\log_b(ca) + 1} + \frac{1}{\log_c(ab) + 1}\)
\(= \frac{1}{=log_a{bc} + \log_a a} + \frac{1}{\log_b(ca) + \log_b b} + \frac{1}{\log_c(ab) + \log_c c}\)
\(= \frac{1}{\log_a(abc)} + \frac{1}{\log_b(abc)} + \frac{1}{\log_c(abc)}\)
\(= \log_{abc}a + \log_{abc}b + \log_{abc}c = \log{abc}abc = 1\)
We have to prove that \(\frac{1}{1 + \log_b a + \log_b c} + \frac{1}{1 + \log_c a + \log_c b} + \frac{1}{1 + \log_a b + \log_a c} = 1\)
L.H.S. \(= \frac{1}{\log_b b + \log_b a + \log_b c} + \frac{1}{\log_c c + \log_c a + \log_c b} + \frac{1}{\log_a a + \log_a b + \log_a c}\)
\(= \frac{1}{\log_b abc} + \frac{1}{\log_c abc} + \frac{1}{\log_a abc}\)
Like previus exercise above expression evaluates to \(1\)
We have to prove that \(x^{\log y - \log z}y^{\log z - \log x}z^{\log x - \log y} = 1\)
Taking log of both sides
\((\log y - \log z)\log x + (\log z - \log x)\log y + (\log x - \log y)\log z = 0\)
\(\log y\log x - \log z\log x + \log z\log y - \log x\log y + \log x\log z - \log y\log z = 0\)
\(0 = 0\)
We have to prove that \(a^xb^yc^z = 1\)
\(x\log a + y\log b + z\log c = 0\)
Let \(\frac{\log a}{y - z} = \frac{\log b}{z - x} = \frac{\log c}{x - y} = k\)
\(x\log a = k(xy - zx), y\log b = k(yz - xy), z\log c = k(zx - yz)\)
Adding all these
\(x\log a + y\log b + z\log c = k(xy -zx + yz - xy + zx - yz) = 0\)
Let \(\frac{x(y + z - x)}{\log x} = \frac{y(z + x - y)}{\log y} = \frac{z(x + y - z)}{\log z} = \frac{1}{k}\)
\(\log x = kx(y + z - x), \log y = ky(z + x - y), \log z = kz(x + y - z)\)
Let \(y^zz^y = z^xx^z = x^yy^x = c\)
Taking log \(z\log y + y\log z = x\log z + z\log x = y\log x + x\log y = log c\)
\(\Rightarrow zky(z + x - y) + ykz(x + y - z) = xkz(x + y - z) + zkx(y + z - x)\\ = ykx(y + z - x) + xky(x + z - y)\)
\(\Rightarrow yz^2 + xyz - y^2z + xyz + y^2z - z^2y = x^2z + xyz -xz^2 + xyz + xz^2 - x^2z\\ = xy^2 + xyz - x^2y + x^2y + xyz - xy^2\)
\(\Rightarrow 2xyz = 2xyz = 2xyz\)
Let \(\frac{\log a}{b - c} = \frac{\log b}{c - a} = \frac{\log c}{a - b} = k\)
\(\log a = k(b - c), \log b = k(c - a), \log c = k(a - b)\)
We have to prove that \(a^{b + c}b^{c + a}c^{a + b} = 1\)
Taking log of both sides, we get
\((b + c)\log a + (c + a)\log b + (a + b)\log c = 0\)
Substituting the values, we get
\(k(b^2 - c^2) + k(c^2 - a^2) + k(a^2 - b^2) = 0\)
Let \(\frac{\log x}{q - r} = \frac{\log y}{r - p} = \frac{\log z}{p - q} = k\)
\(\log x = k(q - r), \log y = k(r - p), \log z = k(p - q)\)
We have to prove that \(x^{q + r}y^{p + r}z^{p + q} = x^py^qz^r\)
Taking log of both sides, we get
\((q + r)\log x + (p + r)\log y + (p + q)\log z = p\log x + q\log y + r\log z\)
Substituting the value of \(\log x, \log y, \log z\)
\(k(q^2 - r^2) + k(r^2 - p^2) + k(p^2 - q^2) = k(pq - pr + qr - pq + pq - qa)\)
\(0 = 0\)
Given \(y = a^{\frac{1}{1 - \log_a x}}\) and \(z = a^{\frac{1}{1 - \log_a y}}\)
\(z = a^{\frac{1}{1 - \log_a (a^{\frac{1}{1 - \log_a x}})}}\)
\(z = a^{\frac{1}{1 - \frac{1}{1 - \log_a x}}}\)
Taking log of both sides with base \(a,\) we get
\(\log_a z = \frac{1}{1- \frac{1}{1 - \log_a x}}\)
\(\log_a z = \frac{1- \log_a x}{-\log_a x} = 1 - \frac{1}{\log_a x}\)
\(x = a^{\frac{1}{1 - \log_a z}}\)
Given \(f(y) = e^{f(z)}\) and \(z= e^{f(x)},\) where \(f(x) = \frac{1}{1 - \log_e x}\)
\(f(y) = e^{\frac{1}{1 - \log_e z}}\)
\(z = e^{\frac{1}{1- \log_e x}}\)
\(f(y) = e^{\frac{1}{1 - \log_e z}} = e^{\frac{1}{1 - \log_e e^{\frac{1}{1 - \log_e x}}}}\)
\(f(y) = e^{\frac{1}{1 - \frac{1}{1 -\log_e x}}}\)
Following like exercise above
\(x = e^{f(y)}\)
L.H.S. \(= \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \ldots + \frac{1}{\log_{43} n}\)
\(= \log_n 2 + \log_n 3 + \log_n 4 + \ldots + \log_n 43\)
\(=\log_n(2.3.4\ldots 43) = \log_n 43! = \frac{1}{\log_{43!}n} =\) R.H.S.
L.H.S. \(= 2(\log a + \log a^2 + \log a^3 + \ldots + \log a^n)\)
\(=2\log a(1 + 2 + 3 + \ldots + n) = 2\log a \frac{n(n + 1)}{2}\)
\(= n(n + 1)\log a\)
We will make use of the fact that positive characteristics of \(n\) of a logarithm means there are \(n + 1\) digits in the number.
Let \(\log y = 12\log 12 = 12\log 2.2.3\)
\(= 12[2*.301 + .477] = 12.96\)
Thus, number of digits is \(13\).
Here we can use the result that number of positive integers having base \(b\) and characteristics \(n\) is \(b^{n + 1}- b^n\)
Thus, number of integers with base \(3\) and characteristics \(2\) is \(3^3 - 3^2 = 18.\)
L.H.S. \(= \log_a x\log_b y\)
\(= \frac{\log x}{\log a}\frac{\log y}{\log b}\)
\(= \frac{\log x}{\log b}\frac{\log y}{\log a}\)
\(\log_bx\log_ay =\) R.H.S.
Given, \(a, b, c\) are in G.P.
\(\frac{b}{a} = \frac{c}{b}\)
\(\log_x \frac{b}{a} = \log_x \frac{c}{b}\)
\(\log_x b - \log_x a = \log_x c - \log_x b\)
\(\log_xa, \log_xb, \log_xc\) are in A.P.
\(\log_a x, \log_b x, \log_c x\) are in H.P.
Let \(y = (0.0504)^{10}\)
\(\log_{10}y = 10\log_{10}(.0504) = 10\log_{10}(504*10^{-4})\)
\(= -10\log_{10}[-4 + \log(2^3.3^2.7)]\)
\(= -12.98\)
Thus, chraracteristics is \(-13\)
Therefore, number of zeroes after decimal and first significant digit \(= 12\)