## Solved Examples in Logarithms

#### Algebra > Logarithms > Solved Examples

**13.Solved Examples in Logarithms: **
Now let us solve a few number of problems on logarithms to apply all of the formulas and concepts learned in this lesson:

**1.Solve the following for x**

1. log 10[ (log _{3} (log _{4} 64)]

2. log 5 (log _{6} 36) = log _{x} 4

*Solution1:*
log _{4} 64 = log _{4} 43

= 3 log _{4} 4 {from Law 3. log _{a} (p)n = n log _{a} p }

= 3.1= 3 { log _{4} 4 = 1, since log of every number to base itself is 1}

Set log _{4} 64 = 3 in the given question, we get

log _{10}[ (log _{3} 3)]

= log _{10} 1 { log_{3}3 = 1}

= 0 { because log of 1 to any base is 0}

*Solution2:*
log _{4} (log _{6} 36) = log _{x} 4

log _{6} 36

= log _{6} 6^{²} { log _{a} (p)^{n} = n log _{a} p}

= 2log _{6} 6 { log 66=1, since log of every number to base itself is 1}

=2

set 2 in log _{6} 36 in the question, we get

log _{4} (log _{6} 36) = log _{4} 2

set this value in the question and we have

log _{4} 2 = log _{x} 4……………………….(1)

now log _{4}2 = log _{2²} 2

= ½ (log _{2}2) = ½ (1) = ½

Put this value ½ in (1), we get

½ = log _{x} 4

x_{1/2} = 4

squaring of both sides, we get

x = 4² = 16

**2. Solve log **_{2} (-p_{2} +10p) = 4

*Solution:*
From log definition, we have
if log _{a} x = n, then a^{n} = x

so, if log 2 (-p2 +10p) = 4, then

-p^{2} +10p = 2^{4}

-p^{2} +10p = 16

p^{2} -10p +16 = 0

we now have a __quadratic equation__ in p.

we must solve this quadratic equation in p with the __method of factorization__
p^{2} - 8p – 2p + 16 = 0

p(p – 8 )- 2 (p – 8) = 0

(p – 2)(p – 8) = 0

So, p – 2 = 0 or p – 8 = 0

So, p = 2 or p = 8

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