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by anshumishra » Fri Jan 07, 2011 3:02 pm
nickhar130 wrote:can someone tell me step by step how to solve this

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

solve for n?
5^21*4^11 = 2*10^n
=> 5^21*2^22 = 2*10^n
=> (5*2)^22 = 5*2*10^n {Multiplied both sides by 5}
=> 10^22 = 10^(n+1)
=> n+1 = 22
=> n = 21.
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Anshu

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by nickhar130 » Fri Jan 07, 2011 3:09 pm
what happened to the 5^21?
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by anshumishra » Fri Jan 07, 2011 3:13 pm
nickhar130 wrote:what happened to the 5^21?
5^21*2^22 = 2*10^n
Multiply both sides by 5
5*5^21*2^22 = 5*2*10^n
=> 5^22*2^22 = 10*10^n
=> (5*2)^22 = 10^(n+1)
=> 10^22 = 10^(n+1)
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by nickhar130 » Fri Jan 07, 2011 3:22 pm
ah.. i see. So what is my goal whenever i encounter these type of questions? Do i make the base numbers similiar so that we can solve the exponents.
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by anshumishra » Fri Jan 07, 2011 3:26 pm
nickhar130 wrote:ah.. i see. So what is my goal whenever i encounter these type of questions? Do i make the base numbers similiar so that we can solve the exponents.
Yes.
In fact you can solve it in an even easier way :

5^21 * 4^11 = 2 * 10^n
=> 5^21*2^22 = 2*10^n
Now divide both sides by 2
=> 5^21*2^21 = 10^n
=> (5*2)^21 = 10^n
=> 10^21 = 10^n
=> n = 21

So, basically you should play with the powers in way so that it is easier to relate them. Here, my aim has been to keep both sides as a power of 10.
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