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Conditions for the Base

Prove that: ...

Question

Prove that:
$log (1 + 2 + 3) = log 1 + log 2 + log 3$ .

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Solution

We have,
$log (1 + 2 + 3)$
$= log (6)$
$= log (1.2.3)$
$= log 1 + log 2 + log 3$ .

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Prove that

$(i) l o g 12 = l o g 3 + l o g 4$

$(i i) l o g 50 = l o g 2 + 2 l o g 5$

$(i i i) l o g (1 + 2 + 3) = l o g 1 + l o g 2 + l o g 3$

{log}_{3} ({log}_{2} x) + {log}_{1 / 3} ({log}_{1 / 2} x) = 1

x

log 2 = a

log 3 = b

[log (1) + log (1 + 3) + log (1 + 3 + 5) + . . . . + . . . . + log (1 + 3 + 5 + 7 + . . . . + 19)]

- 2 [log 1 + log 2 + log 3 + . . . . log 7] = p + q a + r b

p, q, r,

p + 2 q + 3 r

10

log 2 = a a n d log 3 = b

[log (1) + log (1 + 3) + log (1 + 3 + 5) + . . . + . . . + log (1 + 3 + 5 + 7 + . . . + 19)]

2 [log 1 + log 2 + log 3 + . . . + log 7] = p + q a + r b

p, q, r

p + 2 q + 3 r

10

log \frac{162}{343} + 2 log \frac{7}{9} - log \frac{1}{7} = log 2

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