Prove that- log2e-log4e-log8e-1-

Now, log_2e log_4e+log_8e ......∞ =log_2e log_2^2e+log_2^3e ......∞ =log_2e(1 12+13 14+......∞) =log_2 e.log_e 2 [ Since log_e2=1 12+13 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Logarithms

Prove that: ...

Question

Prove that:
${log}_{2} e - {log}_{4} e + {log}_{8} e - . . . . . . \infty = 1$ .

Open in App

Solution

Suggest Corrections

Similar questions

\frac{1}{{log}_{2} 4} + \frac{1}{{log}_{4} 4} + \frac{1}{{log}_{8} 4} + . . . . . . . . + \frac{1}{{log}_{2^{n}} 4}

\frac{1}{{log}_{2} 4} + \frac{1}{{log}_{4} 4} + \frac{1}{{log}_{8} 4} + . . . . . + \frac{1}{{log}_{2 n} 4} .

{log}_{3} 2 \times {log}_{4} 3 \times {log}_{5} 4 \times {log}_{6} 5 \times {log}_{7} 6 \times {log}_{8} 7 = \frac{1}{3}

{log}_{4} (x + 4) + {log}_{4} 8 = 2

{log}_{6} (x + 4) - {log}_{6} (x - 1) = 1

{log}_{2} x + {log}_{4} x + {log}_{8} x = \frac{11}{6}

{log}_{4} (8 {log}_{2} x) = 2

{log}_{10} 5 + {log}_{10} (5 x + 1) = {log}_{10} (x + 5) + 1

4 {log}_{2} x - {log}_{2} 5 = {log}_{2} 125

{log}_{3} 25 + {log}_{3} x = 3 {log}_{3} 5

{log}_{3} (\sqrt{5 x - 2}) - \frac{1}{2} = {log}_{3} (\sqrt{x + 4})

{log}_{8} a + {log}_{4} b^{2} = 5, {log}_{8} b + {log}_{4} a^{2} = 7

a b

Join BYJU'S Learning Program