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If 2^8x=640000 : Multiple-choice Questions — Select One Answer Choice

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Re: Any short-cut way to solve this problem? [#permalink]
Carcass wrote:
\({2^2}^{3x} = 2^{3^2} \times 5^4\)

Originally posted by huda on 24 Jun 2019, 04:27.
Last edited by huda on 24 Jun 2019, 05:10, edited 3 times in total.
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Re: Any short-cut way to solve this problem? [#permalink]
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Sorry. I was editing the post. It was not completed.

See now :)
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Re: Any short-cut way to solve this problem? [#permalink]
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Hi,

which step you didnt get ??
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Re: Any short-cut way to solve this problem? [#permalink]
Carcass wrote:
Hi,

which step you didnt get ??


4th line RHS. How did you get 5th line RHS from 4th line RHS?
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Re: Any short-cut way to solve this problem? [#permalink]
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it is an exponent tule

\(2^{6x} - 2^6\)

Same base, then you can subtract the exponents

\(2^{6x-6}\)
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Re: Any short-cut way to solve this problem? [#permalink]
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Carcass wrote:
\({2^2}^{3x} = 2^{3^2} \times 5^4\)


shouldnt the x next to the 2 on the LHS?

Because the way I am reading the first line, is 2 to the power of 2 to the power of 3 x, and when you power it.

it would be 2 to the power of 8 to the power of x?


I dont understand why the x was carried over to the higher power than brought down?
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Re: If 2^8x=640000 [#permalink]
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It is not brought down.

it is a number raised to the power and this raised to the power.

it is 2 raised to the two and the latter raised to 3x

\({2^2}^{3x}\)

Then you multiply the exponents and the result is \(2^{6x}\)
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Re: Any short-cut way to solve this problem? [#permalink]
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Carcass wrote:
\({2^2}^{3x} = 2^{3^2} \times 5^4\)

\(2^{6x} = 2^6 \times 5^4\)



I guess, I dont know where the 8 went?

In the answer above, is the LHS or the RHS the \(2^{8x}\)\(=640000\)?


edit:
In the problem, it is 8x, and x is multiplied by 8. I understand that \(2^{8}\) is like saying \({2^2}^{3}\). I am having a hard time getting why the x is next to the three, not the second two.


I also understand if the \({2^{{(2x)}^{3}}\), then the x will be cubed

Originally posted by h8gre on 24 Jun 2019, 16:29.
Last edited by h8gre on 24 Jun 2019, 16:45, edited 1 time in total.
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Re: Any short-cut way to solve this problem? [#permalink]
2
Carcass wrote:
\({2^2}^{3x} = 2^{3^2} \times 5^4\)

\(2^{6x} = 2^6 \times 5^4\)

\(2^{6x-6} = 5^4\)

\(\frac{2^{6x-6}}{2^3}= \frac{5^4}{2^3}\)

On the LHS simplified we do have the form we are looking for

\(2^{2x-2} = 5^4\)

Only A fits the RHS

More than this is impossible. manipulation and a touch of educating guess


I mean,
How did you get this \(5^4\) from this \(\frac{5^4}{2^3}\)?
or from \(\frac{2^{6x-6}}{2^3}= \frac{5^4}{2^3}\) to \(2^{2x-2} = 5^4\)? Need overall explanation because it not easy for me to understand.
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Re: If 2^8x=640000 [#permalink]
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Yes you are right. I made a mistake to think it was raised to x and not multiplied by x

Ok attack from another perspective

1) \(2^{8x} = 2^6 \times 10^4\)

2) \(2^{8x} = 2^6 2^4 \times 5^4\)

OR

\(\frac{2^{8x}}{2^8}= \frac{2^8 \times 2^2 \times 5^4}{2^8}\)

\(2^{8x-8} = 2^2 \times 5^4\)

Now clearly all the exponents can be divided by four

\(2^{2x-2} = 2^{\frac{1}{2}} \times 5\)

\(2^{2x-2} = 5 \sqrt{2}\)
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Re: If 2^8x=640000 [#permalink]
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Carcass wrote:
Yes you are right. I made a mistake to think it was raised to x and not multiplied by x

Ok attack from another perspective

1) \(2^{8x} = 2^6 \times 10^4\)

2) \(2^{8x} = 2^6 2^4 \times 5^4\)

OR

\(\frac{2^{8x}}{2^8}= \frac{2^8 \times 2^2 \times 5^4}{2^8}\)

\(2^{8x-8} = 2^2 \times 5^4\)

Now clearly all the exponents can be divided by four

\(2^{2x-2} = 2^{\frac{1}{2}} \times 5\)

\(2^{2x-2} = 5 \sqrt{2}\)


Fully understand, Thanks. :) :thanks :gl :done
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Re: If 2^8x=640000 [#permalink]
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Carcass wrote:

\(2^{8x-8} = 2^2 \times 5^4\)

Now clearly all the exponents can be divided by four

\(2^{2x-2} = 2^{\frac{1}{2}} \times 5\)

\(2^{2x-2} = 5 \sqrt{2}\)



Can we divide the exponents by 4 irrespective of the base number?

Please can you provide an explanation. ( I have never came acrossed this rule :? )
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Re: If 2^8x=640000 [#permalink]
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pranab01 wrote:
Carcass wrote:

\(2^{8x-8} = 2^2 \times 5^4\)

Now clearly all the exponents can be divided by four

\(2^{2x-2} = 2^{\frac{1}{2}} \times 5\)

\(2^{2x-2} = 5 \sqrt{2}\)



Can we divide the exponents by 4 irrespective of the base number?

Please can you provide an explanation. ( I have never came acrossed this rule :? )


Exactly. That was my question too.

The final equation stands like this..
\(2^{8x-10} = 5^{4}\)
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Re: If 2^8x=640000 [#permalink]
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First of all the exponent is 8x-8 and not 8x-10

Secondly, I went down without some obvious step but indeed yes.

It is a simplification/manipulation.

regards
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Re: If 2^8x=640000 [#permalink]
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I had been waiting for you Sir and your insight.

Thank you
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Re: If 2^8x=640000 [#permalink]
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Carcass wrote:
First of all the exponent is 8x-8 and not 8x-10

Secondly, I went down without some obvious step but indeed yes.

It is a simplification/manipulation.

regards


Got that,

Just trying out to simplify

\(2^{8x} = 640000\)

or \(2^{8x} = 2^6 * 2^4 * 5^4 = 2^{10} * 5^4\) ( 640000 = we know, 2^6 = 64 and now there are 4 zeros after 64, that means it will compiled with = 2^4 * 5^4 , since 2*5 =10)

Now taking square root on both sides

\(2^{4x} = 2^5 * 5^2\)

Dividing by \(2^4\)on both sides

\(\frac{2^{4x}}{{2^4}} = \frac{{2^5*5^2}}{{2^4}}\)

\(\frac{2^{4x}}{{2^4}}= 2 * 5^2\)

Taking square root again,

\(\frac{2^{2x}}{{2^2}} = 5 \sqrt{2}\)

or \(2^{2x - 2} = 5 \sqrt{2}\)
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Re: If 2^8x=640000 [#permalink]
Carcass wrote:
First of all the exponent is 8x-8 and not 8x-10

Secondly, I went down without some obvious step but indeed yes.

It is a simplification/manipulation.

regards


Dude. I am just quoting what you said before.

Now clearly all the exponents can be divided by four


\(2^{8x-8} = 2^{2} * 5^{4}\)

Dividing by 4 will give you

\(2 ^{8x-10} = 5^{4}\)

Am I missing anything? :shock:

Originally posted by harishsridharan on 25 Jun 2019, 07:01.
Last edited by harishsridharan on 25 Jun 2019, 07:05, edited 1 time in total.
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