Question
Given that log5 3=0.683 and log5 7=1.209.without using calculator,evaluate
(a) log5 1.4
b)log7 75
c) log3 125
(a) log5 1.4
b)log7 75
c) log3 125
Answers
Did you notice that 1.4 = 7/5 ?
log<sub>5</sub> 1.4
= log<sub>5</sub> (7/5)
= log<sub>5</sub> 7 - log<sub>5</sub> 5
= 1.209 - 1
= .209
A property of logs that comes in handy here is
log<sub>a</sub> b = 1/log<sub>b</sub> a
b) log<sub>7</sub> 75
= log<sub>7</sub> (3*5^2)
=log<sub>7</sub> 3 + 2log<sub>7</sub> 5
= (log<sub>5</sub> 3) / (log<sub>5</sub> 7) + 2/log<sub>5</sub> 7)
= .683/1.209 + 2/1.209
= 2.683/1.209
= 2683/1209 ----> At this point I used Calculator to get appr 2.219
c) log<sub>3</sub> 125
= log<sub>3</sub> 5^3
= 3 log<sub>3</sub> 5
= 3 (1/log<sub>5</sub> 3)
= 3/.683
log<sub>5</sub> 1.4
= log<sub>5</sub> (7/5)
= log<sub>5</sub> 7 - log<sub>5</sub> 5
= 1.209 - 1
= .209
A property of logs that comes in handy here is
log<sub>a</sub> b = 1/log<sub>b</sub> a
b) log<sub>7</sub> 75
= log<sub>7</sub> (3*5^2)
=log<sub>7</sub> 3 + 2log<sub>7</sub> 5
= (log<sub>5</sub> 3) / (log<sub>5</sub> 7) + 2/log<sub>5</sub> 7)
= .683/1.209 + 2/1.209
= 2.683/1.209
= 2683/1209 ----> At this point I used Calculator to get appr 2.219
c) log<sub>3</sub> 125
= log<sub>3</sub> 5^3
= 3 log<sub>3</sub> 5
= 3 (1/log<sub>5</sub> 3)
= 3/.683
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