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

If
\ log3 M = a{1} + b{1}
and
\ log5 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.....

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Comments

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.

Pi Han Goh - 8 years, 1 month ago

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

A Former Brilliant Member - 8 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$