Help with a basic Exponents concept

Question

Hi,

Can someone please help me with one of the exponent concepts?

Say we have 2 integers a,b

Lets assume this equation

a^x= b*a^13.

Now if we are required to find the value of x, do we need to get rid of b? of can we ignore b and say x=13 directly. Basically my understanding isnt clear even after reading from Manhattan.

Thanks a lot

suyash23n · Answer

Hi. in the given equation ,b can be ignored only when we are dealing with prime numbers (say a & b, both are prime) and you can directly get the answer. However, with non prime numbers, you can not ignore b. In such cases, you must factorize a & b to prime numbers and then proceed.

Hope it Helps!

tchang · Answer

Hi,

I've been working on exponent rules, so here's what I think is happening here...

If we have a^x = b(a^13) and a, b are integers, then we need to know more about a and b to solve for x.

a^x is just a bunch of a's multiplied

eg a^3 = a(a)(a)

So if a^x = b(a^13)

then we know that "b" must equal a^(something); it's the only way that the left side of the equation can be a^x.

If b = 1, then a^x = 1(a^13) so x = 13
If b = 2 then a must also = 2, then we'd have 2^x = 2(2^13) so x = 14
If b = 3, then a must also = 3, then we'd have 2^x = 3(3^13) so x = 14
If b = 9, then a could be 3 or 9
If b = 9 and a = 3, then we'd have 3^x = 9(3^13) = 3^2(3^13) so x = 15
etc.

t1000

KarishmaB · Answer

No. You cannot ignore b. 
e.g.
 5^x = b*5^{13} 
Can you say here that x = 13? Absolutely not. Until and unless b is 1, we can say for sure that x is NOT 13.

Say b = 25.
 5^x = 25*5^{13} 
 5^x = 5^2*5^{13} 
x = 15

Say b = 30.
 5^x = 30*5^{13} 
 5^x = 6*5^{14} 
 5^{x-14} = 6 
Now you take log on both sides to figure out the value of x which will not be an integer.

x = 14 + log6/log5

When do we equate powers on left hand side to powers on right hand side independently? When dealing with prime bases and integer powers. Say

3^a * 5^{10} = 15^2 * 5^b  (given a and b are integers)
 3^a * 5^{10} = 3^2*5^2 * 5^b 
 3^a * 5^{10} = 3^2 * 5^{2 + b}

Now we have prime bases on both sides and powers must be integers so we can equate them independently.
a = 2
2 + b = 10

If the powers needn't be integers, a and b can take any values. Say a = 1,
 3 * 5^{10} = 3^2 * 5^{2 + b}

5^{2 + b) = \frac{5^{10}}{3}

Now you will take log on both sides and get a decimal value for b.

Help with a basic Exponents concept

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