I need to know the concept of (a^m)*(b^n). for eg (5^21)*(4^11) = 2*10^n. where n=21. How do we get value of n

Question

Something related to exponents

Answer 1

these answers are of no use

simply write 5=10/2

[(10)^21] / [(2)^21] * {(2)^11 * (2)^11}

= (10)^21 * (2)^1

u know how i got 2^1 by just subtaction (2)^{22-21}

isn't it easy

Answer 2

The approach to the question depend on what you have studied so far. This a very simple question and can be tackled using the properties of powers. However, if you are doing A-levels, then, you are likely to get harder questions of the same form and expected to use logarithms and exponentials.

If you are reading at year 11 or below, here is a way of solving this question:
We have (5^21)*(4^11) = 2*10^n.
But, (5^21)*(4^11) = (5^21)*((2*2)^11)
= (5^21)*((2^2)^11)
= (5^21)*(2^22)
= (5^21)*(2^21)*2
= 2*(5*2)^21
= 2*10^21
This is in the required form and therefore, n = 21.
Note: In solving this question, I have used the fact that
(a^n)*(b^n) = (a*b)^n = (ab)^n

[If you are not doing A-levels, then please ignore the answer below.]

Here is another way of approaching the answer.
We have (5^21)*(4^11) = 2*10^n.
Then, using the properties of logarithms,
log((5^21)*(4^11)) = log(2*10^n)
log(5^21) + log(4^11) = log(2) + log(10^n)
21*log(5) + 11*log(4) = log(2) + n*log(10)
21*log(5) + 11*log(2*2) = log(2) + n*log(2*5)
21*log(5) + 22*log(2) = log(2) + n*log(2) + n*log(5)
21*log(5) + 22*log(2) = (n+1)*log(2) + n*log(5)
Now, equating the coefficients gives: n = 21 as requires.

Note: In solving this question, I have used the following properties of logarithms
log(a*b) = log(a) + log(b)
log(a^b) = b*log(a)

Answer 3

Remember the rules that (a^b)(a^c)=a^(b+c) and (ab)^c = (a^c)(b^c). We can use these rules in reverse to factor out some of the numbers:

(5^21)(4^11) = 2(10^n)
(5^21)(2*2)^11 = 2(10^n)
(5^21) (2^11)(2^11) = 2(10^n)
(5^21) 2^22 = 2(10^n)
(5^21) 2^21 = (10^n)
(5*2)^21 = (10^n)
10^21 = 10^n
n = 21

If you've already learned about logarthimns (and if not, ignore the following), you can solve it with logs too:

(5^21)(4^11) = 2*10^n
log [(5^21)(4^11)] = log (2*10^n)
log (5^21) + log(4^11) = log2 + log(10^n)
log (5^21) + log (4^11) - log(2) = n
21*log (5) + 11*log (4) - log(2) = n
21*log (5) + 22*log (2) - log(2) = n
21*log (5) + 21*log (2) = n
21(log (5) + log (2)) = n
21(log (10)) = n
n = 21

Answer 4

(5^21)(4^11)
= (5^21)[(2^2)^11]
= (5^21)(2^22)
= (5^21)(2)(2^21)
= 2 [(5)(2)]^21
= 2 (10^21) = 2*10^n
n=21

You can do it by taking log of both the side of equation
21log5+11Log4 = log2+nlog10
21log5+22log2 = log2 + nlog10
21(log5+log2) = nlog10
21log10 = nlog10
hence n = 21

Answer 5

(5^21)(4^11)
= (5^21)[(2^2)^11]
= (5^21)(2^22)
= (5^21)(2)(2^21)
= 2 [(5)(2)]^21
= 2 (10^21)

Answer 6

(5^21)*(4^11) = 2*10^n
(5^21)*(2^22) = 2*10^n
(5^21)(2^21) = 10^n
(5^21)(2^21) = (5*2)^n = 5^n2^n
Hence n = 21

Answer 7

(5^21)*(4^11) = 2*10^n

n
= log[(5^21)*(2^22)/2]/log10
= 21

Answer 8

The only thing I can tell you is that if a = b, then the product of a and b is the base (a or b) raised to the sum of m and n...this only holds for expressions with the same base.