If

\ log*3 M = a*{1} + b*{1}
and
\ log*5 M = a

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.....

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestAnswer 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.

Log in to reply

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

Log in to reply