If $4 = 10^{2m}$ and $9 = 10^{2n}$, then $0 \\cdot 15$ equals to :<\/p>\n
[amp_mcq option1=”$10^{2m-2n}$” option2=”$10^{m+n-1}$” option3=”$10^{n-m-1}$” option4=”$10^{m-n-1}$” correct=”option3″]<\/p>\n
From the first equation:
\n$4 = (10^m)^2$
\nTaking the square root of both sides:
\n$\\sqrt{4} = \\sqrt{(10^m)^2}$
\n$2 = 10^m$ (Assuming $10^m$ is positive, which is true for real $m$ as $10^x$ is always positive)<\/p>\n
From the second equation:
\n$9 = (10^n)^2$
\nTaking the square root of both sides:
\n$\\sqrt{9} = \\sqrt{(10^n)^2}$
\n$3 = 10^n$ (Assuming $10^n$ is positive)<\/p>\n
We need to express $0.15$ using $10^m$ and $10^n$.
\n$0.15 = \\frac{15}{100}$
\nWe know $10^n = 3$. Let’s try to express 15 and 100 using 2, 3, and powers of 10.
\n$0.15 = \\frac{3 \\times 5}{10^2}$
\nWe have 3 ($10^n$). We need to get 5 and relate it to $10^m$ and $10^n$.
\n$10 = 10^1$. We know $10^m=2$ and $10^n=3$. $10 = 2 \\times 5 = 10^m \\times 5$. So, $5 = 10 \/ 10^m = 10^1 \/ 10^m = 10^{1-m}$.<\/p>\n
Substitute $3 = 10^n$ and $5 = 10^{1-m}$ into the expression for 0.15:
\n$0.15 = \\frac{10^n \\times 10^{1-m}}{10^2}$
\nUsing exponent rules ($a^x \\times a^y = a^{x+y}$ and $a^x \/ a^y = a^{x-y}$):
\n$0.15 = 10^{n + (1-m) – 2}$
\n$0.15 = 10^{n + 1 – m – 2}$
\n$0.15 = 10^{n – m – 1}$<\/p>\n
Let’s check this against the options. Option C is $10^{n-m-1}$. This matches our result.<\/p>\n