1. ## INDICES PROBLEM SOLVING

X = 4 × 10m Y = 2 × 10n
Where m and n are integers.
XY= 8 × 102 and x/y = 2 × 108
Find the value of m and the value of n

Originally Posted by abey_27
X = 4 × 10m Y = 2 × 10n
Where m and n are integers.

XY= 8 × 102 and x/y = 2 × 108
Find the value of m and the value of n
$XY = 8 \times 102$

$(4 \times 10m)(2 \times 10n) = 8 \times 102$

$8 \times 100 \times mn = 8 \times 102$

$mn = \frac{102}{100}$ .........eqn(1)

$\frac{X}{Y} = 2 \times 108$

$\frac{4 \times 10m}{2 \times 10n} = 2 \times 108$

$\frac{2m}{n} = 2 \times 108$

$\frac{m}{n} =108$

$m = 108 n$ ..............eqn(2)

Put this value of m in eqn (1),

$(108n)n = \frac{102}{100}$

$108n^2 = \frac{102}{100}$

$n^2 = \frac{102}{100 \times 108}$

$n= \sqrt{\frac{102}{100 \times 108}}$, now, simplify to get integer value

put this value of n in eqn(1),

$m = 108 \times \sqrt{\frac{102}{100 \times 108}}$

$m = \sqrt{\frac{108 \times 102}{100}}$, now, simplify to get integer value

3. sorry shyam, i forgot to put the power indices symbol on, could you possibly do the Q again, much appreciated thanks

X = 4 × 10^m Y = 2 × 10^n
Where m and n are integers.
XY= 8 × 10^2 and x/y = 2 × 10^8
Find the value of m and the value of n

4. ## Solution involves simultaneous equations.

x=4X10^m and y=2X10^n

xy=8X10^2

hence (4X10^m)(2X10^n) = 8X10^2
so 8X10^(m+n) = 8X10^2
simplified: 10^(m+n) = 10^2 (divide by 8)
hence m+n = 2

x/y=2X10^8

hence (4X10^m)/(2X10^n) = 2X10^8
so 2X10^(m-n) = 2X10^8
simplified: 10^(m-n) = 10^8 (divide by 2)
hence m-n = 8

now m+n = 2
+ m-n = 8
_____________

2m = 10 (since n-n = 0)

hence m = 5 (since 2X5 = 10)

substitute into m+n = 2
5+n = 2
n = 2 - 5
n = -3

so: m=5 and n=-3

Lovely question. Hope this helps. Cheers.

5. thank you so much, could you possibly try the other indice question as well, highly appreciated

6. Originally Posted by abey_27
X = 4 × 10^m Y = 2 × 10^n
Where m and n are integers.
XY= 8 × 10^2 and x/y = 2 × 10^8
Find the value of m and the value of n
$XY = 8 \times 10^{n+m}$ and $X/Y = 2 \times 10^{n-m}$.

This means $n+m = 2$ and $n-m=8$.
Can you solve these equations?

7. yea, thanks you for your help. I have just posted another algebra Q, if you have time, could you possibly help me on that one. thank you again

8. Originally Posted by abey_27
sorry shyam, i forgot to put the power indices symbol on, could you possibly do the Q again, much appreciated thanks

X = 4 × 10^m Y = 2 × 10^n
Where m and n are integers.
XY= 8 × 10^2 and x/y = 2 × 10^8
Find the value of m and the value of n

$XY = 8 \times 10^2$

$(4 \times 10^m)(2 \times 10^n) = 8 \times 10^2$

$8 \times 10^{m+n} = 8 \times 10^2$

$10^{m+n} = 10^2$

$m+n = 2$ .........eqn(1)

Now,

$\frac{X}{Y} = 2 \times 10^8$

$\frac{4 \times 10^m}{2 \times 10^n} = 2 \times 10^8$

$2\times 10^{m-n}= 2 \times 10^8$

$10^{m-n}=10^8$

$m-n=8$ ................eqn(2)

$m+n = 2$
$m-n=8$
-----------------------------------
2m = 10
-----------------------------------

m = 5

Put this value of m in eqn(1),

5 + n = 2

n = - 3

so, m = 5 and n = -3

Did you get it now???