Logarithmic Expressions: Very Difficult Problems with Solutions

Solution:
[tex]\log_{2}3\cdot \log_{3}4 = \log_{2}4[/tex]
[tex]\log_{2}4\cdot \log_{4}5 = \log_{2}5[/tex]
Using [tex]\log_a b\cdot \log_b c = \log_a c[/tex]:
[tex]\log_{2}5\cdot \log_{5}8 = \log_{2}8 = 3[/tex]

Answer: 3

Solution:
[tex]\frac{1}{\log_{2}6} = \log_{6}2[/tex]
[tex]\frac{1}{\log_{3}6} = \log_{6}3[/tex]
Using [tex]\log_a x + \log_a y = \log_a(xy)[/tex]:
[tex]\log_{6}2 + \log_{6}3 = \log_{6}6 = 1[/tex]

Answer: 1

Solution:
[tex]\sqrt{2} = 2^{\frac{1}{2}}[/tex]
[tex]\sqrt[3]{4} = \sqrt[3]{2^2} = 2^{\frac{2}{3}}[/tex]
Using [tex]\log_{a^n}a^m = \frac{m}{n}[/tex]:
[tex]\log_{\sqrt{2}}\sqrt[3]{4} = \log_{2^{\frac{1}{2}}}2^{\frac{2}{3}} = \frac{\frac{2}{3}}{\frac{1}{2}} = \frac{4}{3}[/tex]

Answer: [tex]\frac{4}{3}[/tex]

Solution:
[tex]5 = \frac{10}{2}[/tex]
Using [tex]\log_a\frac{x}{y} = \log_a x - \log_a y[/tex]:
[tex]\log_{10}5 = \log_{10}\frac{10}{2} = \log_{10}10 - \log_{10}2 = 1 - a[/tex]

Answer: [tex]1-a[/tex]

Solution:
[tex]8 = 2^3[/tex]
[tex]8^{\log_{2}5} = (2^3)^{\log_{2}5} = (2^{\log_{2}5})^3[/tex]
Using [tex]a^{\log_a b} = b[/tex]:
[tex](2^{\log_{2}5})^3 = 5^3 = 125[/tex]

Answer: 125

Solution:
[tex]\log_{2}65536 = \log_{2}2^{16} = 16[/tex]
[tex]\log_{2}16 = \log_{2}2^4 = 4[/tex]
[tex]\log_{2}(\log_{2}(\log_{2}65536)) = \log_{2}4 = 2[/tex]

Answer: 2

Solution:
[tex]a>0,\; a\neq 1,\; b>0,\; b\neq 1[/tex]
Using [tex]\log_a b = \frac{1}{\log_b a}[/tex]:
[tex]\log_a b\cdot \log_b a = \frac{1}{\log_b a}\cdot \log_b a = 1[/tex]

Answer: 1

Solution:
[tex]2\log_{10}5 = \log_{10}5^2 = \log_{10}25[/tex]
Using [tex]\log_a x + \log_a y = \log_a(xy)[/tex]:
[tex]\log_{10}4 + \log_{10}25 = \log_{10}100 = 2[/tex]

Answer: 2

Solution:
[tex]\sqrt[4]{27} = \sqrt[4]{3^3} = 3^{\frac{3}{4}}[/tex]
Using [tex]\log_a a^m = m[/tex]:
[tex]\log_{3}\sqrt[4]{27} = \log_{3}3^{\frac{3}{4}} = \frac{3}{4}[/tex]

Answer: [tex]\frac{3}{4}[/tex]

Solution:
[tex]32 = 2^5[/tex]
Using [tex]\frac{\log_c b}{\log_c a} = \log_a b[/tex]:
[tex]\frac{\log_{5}2}{\log_{5}32} = \log_{32}2 = \log_{2^5}2 = \frac{1}{5}[/tex]

Answer: [tex]\frac{1}{5}[/tex]

Solution:
[tex]\log_{4}9 = \frac{\log_{2}9}{\log_{2}4} = \frac{2\log_{2}3}{2} = \log_{2}3[/tex]
Using [tex]a^{\log_a b} = b[/tex]:
[tex]2^{\log_{4}9} = 2^{\log_{2}3} = 3[/tex]

Answer: 3

Solution:
[tex]\log_{8}9 = \frac{\log_{2}9}{\log_{2}8} = \frac{\log_{2}3^2}{3} = \frac{2\log_{2}3}{3}[/tex]
[tex]\log_{8}9 = \frac{2b}{3}[/tex]

Answer: [tex]\frac{2b}{3}[/tex]

Solution:
[tex]\log_{3}2\cdot \log_{4}3 = \log_{4}2[/tex]
[tex]\log_{4}2\cdot \log_{5}4 = \log_{5}2[/tex]
[tex]\log_{5}2\cdot \log_{6}5 = \log_{6}2[/tex]
[tex]\log_{6}2\cdot \log_{7}6 = \log_{7}2[/tex]
Using [tex]\log_b a\cdot \log_c b = \log_c a[/tex]:
[tex]\log_{7}2\cdot \log_{8}7 = \log_{8}2 = \log_{2^3}2 = \frac{1}{3}[/tex]

Answer: [tex]\frac{1}{3}[/tex]

Solution:
[tex]4^{1+\log_{2}3} = 4\cdot 4^{\log_{2}3}[/tex]
[tex]4^{\log_{2}3} = (2^2)^{\log_{2}3} = (2^{\log_{2}3})^2 = 3^2 = 9[/tex]
[tex]4^{1+\log_{2}3} = 4\cdot 9 = 36[/tex]

Answer: 36

Solution:
[tex]\log_{25}27 = \frac{\log_{5}27}{\log_{5}25} = \frac{3\log_{5}3}{2}[/tex]
Using [tex]\log_a b\cdot \log_b a = 1[/tex]:
[tex]\log_{3}5\cdot \log_{25}27 = \frac{3}{2}\cdot \log_{3}5\cdot \log_{5}3 = \frac{3}{2}[/tex]

Answer: [tex]\frac{3}{2}[/tex]

Solution:
[tex](a+b)^2=4ab[/tex]
We know that [tex](a+b)^2 = a^2 + b^2 + 2ab[/tex]
[tex]a^2 + b^2 + 2ab = 4ab[/tex]
[tex]a^2+b^2=4ab-2ab[/tex]
So [tex]a^2+b^2=2ab[/tex], i.e. [tex](a-b)^2=0[/tex] and [tex]a=b[/tex] (with [tex]a>0[/tex] and [tex]b>0[/tex], otherwise the logarithms are not defined)

Then [tex]\log(a^2+b^2)=\log(2ab)=\log 2+\log(a)+\log(b)[/tex]
but [tex]2\log(a)+2\log(b)=\log(a^2b^2)[/tex]
The two are equal only if [tex]2ab=a^2b^2[/tex], that is only if [tex]ab=2[/tex], which does not follow from the condition.
Counterexample: [tex]a=b=1[/tex] satisfies [tex](a+b)^2=4ab[/tex], but [tex]\log(a^2+b^2)=\log 2\approx 0.301[/tex], while [tex]2\log(a)+2\log(b)=0[/tex].

Answer: False