# Logarithm Question............Help!!!

If
\ log3 M = a{1} + b{1}
and
\ log
5 M = a{2} + b{2}

where
a{1} , a{2} \ in N
and
b{1} , b{2} \ in [0,1).
If a{1} * a{2} = 6 , then find the number of integral values of M.

URGENT.....

Note by Riya Gupta
5 years, 4 months ago

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Answer is $$80 - 27 + 1 = 54$$ with $$M = 27, 28, 29, \ldots , 80$$

Given $$log_3 M = a_1 + b_1$$ and $$log_5 M = a_2 + b_2$$

Where $$a_1 , a_2$$ are natural numbers and $$b_1 , b_2$$ are have values in range $$[0, 1)$$

With $$a_1 \cdot a_2 = 6$$

So at least one of these are true

$$a_1 = 1, a_2 = 6$$,

$$a_1 = 2, a_2 = 3$$,

$$a_1 = 3, a_2 = 2$$,

$$a_1 = 6, a_2 = 1$$

Then $$3^{a_1} \leq 3^{a_1 + b_1} < 3^{a_1 + 1}$$ and $$5^{a_2} \leq 5^{a_2 + b_2} < 5^{a_2 + 1}$$

Or $$3^{a_1} \leq M < 3^{a_1 + 1}$$ and $$5^{a_2} \leq M < 5^{a_2 + 1}$$

Suppose $$a_1 = 6, a_2 = 1$$, Then M must satisfy the two inequalities

$$\Rightarrow 3^6 \leq M < 3^7$$ and $$5^6 \leq M < 5^7$$

which yields no common value

Suppose $$a_1 = 2, a_2 = 3$$, Then M must satisfy the two inequalities

$$\Rightarrow 3^2 \leq M < 3^3$$ and $$5^3 \leq M < 5^4$$

which yields no common value

Suppose $$a_1 = 3, a_2 = 2$$, Then M must satisfy the two inequalities

$$\Rightarrow 3^3 \leq M < 3^4$$ and $$5^2 \leq M < 5^3$$

which yields $$M = 27, 28, 29, \ldots , 80$$

Suppose $$a_1 = 1, a_2 = 6$$, Then M must satisfy the two inequalities

$$\Rightarrow 3^1 \leq M < 3^2$$ and $$5^6 \leq M < 5^7$$

which yields no common value

Thus $$M = 27, 28, 29, \ldots , 80$$ only.

- 5 years, 4 months ago

thanks for the simplified solution...i was really in need of it....:)

- 5 years, 4 months ago