Question

Determine the value of the following :
(i) ㏒₂₅5
(ii) ㏒₈₁3
(iii) ㏒₂$\frac{1}{16}$
(iv) ㏒₇1
(v) ㏒ₓ√x
(vi) ㏒₂521
(vii) ㏒₁₀0.01
(viii) ㏒₂₁₃$\frac{8}{27}$
(ix) 2²+㏒₂³

Answer 1

logₐb = (log b)/(log a)
and log (aⁿ) = n log a
i) log₂₅5
= (log5)/(log25) = (log5)/(log5²) 
= (log5)/2(log5) = 1/2

ii)log₈₁3
=(log3)/(log81) = (log3)/(log3^4)
= (log3)/4(log3) = 1/4

iii) log₂(1/16) 
= log(1/16)/log2
= - log 16/log 2
= - (log 2^4)/log 2 = -4 log2/log2 = -4

iv)㏒₇1
log 1/log 7 = 0 
∵log 1 = 0

v)㏒ₓ√x 
=[ log (x)^(1/2)]/ logx = 1/2 logx/logx = 1/2

vi) actually this is ㏒₂ 512 
log 512/log2 = log 2⁹/log 2 
=9log2/log2 = 9

vii)㏒₁₀0.01
=(㏒10⁻²)/log 10 
= -2 log10/log10 = -2

viii) please recheck your question. or solve in the same way
if any doubt..please say

Answer 2

Logₐb = (log b)/(log a)

and log (aⁿ) = n log a

i) log₂₅5

= (log5)/(log25) = (log5)/(log5²)

= (log5)/2(log5) = 1/2

ii)log₈₁3

=(log3)/(log81) = (log3)/(log3^4)

= (log3)/4(log3) = 1/4

iii) log₂(1/16)

= log(1/16)/log2

= - log 16/log 2

= - (log 2^4)/log 2 = -4 log2/log2 = -4

iv)㏒₇1

log 1/log 7 = 0

∵log 1 = 0

v)㏒ₓ√x

=[ log (x)^(1/2)]/ logx = 1/2 logx/logx = 1/2

vi) actually is ㏒₂ 512

log 512/log2 = log 2⁹/log 2

=9log2/log2 = 9

vii)㏒₁₀0.01

=(㏒10⁻²)/log 10

= -2 log10/log10 = -2