Last visit was: 14 Nov 2024, 12:21 It is currently 14 Nov 2024, 12:21

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
User avatar
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 709 [22]
Given Kudos: 161
Send PM
Most Helpful Community Reply
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12188 [14]
Given Kudos: 136
Send PM
General Discussion
Verbal Expert
Joined: 18 Apr 2015
Posts: 29960
Own Kudos [?]: 36219 [0]
Given Kudos: 25903
Send PM
User avatar
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 709 [0]
Given Kudos: 161
Send PM
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.
Verbal Expert
Joined: 18 Apr 2015
Posts: 29960
Own Kudos [?]: 36219 [0]
Given Kudos: 25903
Send PM
Re: Any short-cut way to solve this problem? [#permalink]
Expert Reply
Sorry. I was editing the post. It was not completed.

See now :)
Verbal Expert
Joined: 18 Apr 2015
Posts: 29960
Own Kudos [?]: 36219 [0]
Given Kudos: 25903
Send PM
Re: Any short-cut way to solve this problem? [#permalink]
Expert Reply
Hi,

which step you didnt get ??
User avatar
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 709 [0]
Given Kudos: 161
Send PM
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?
Verbal Expert
Joined: 18 Apr 2015
Posts: 29960
Own Kudos [?]: 36219 [0]
Given Kudos: 25903
Send PM
Re: Any short-cut way to solve this problem? [#permalink]
Expert Reply
it is an exponent tule

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

Same base, then you can subtract the exponents

\(2^{6x-6}\)
avatar
Intern
Intern
Joined: 24 Jun 2019
Posts: 2
Own Kudos [?]: 2 [0]
Given Kudos: 0
Send PM
Re: Any short-cut way to solve this problem? [#permalink]
1
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?
Attachments

File comment: pocture
sss.PNG
sss.PNG [ 201.13 KiB | Viewed 9044 times ]

Verbal Expert
Joined: 18 Apr 2015
Posts: 29960
Own Kudos [?]: 36219 [0]
Given Kudos: 25903
Send PM
Re: If 2^8x=640000 [#permalink]
Expert Reply
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 \(2^{6x}\)
avatar
Intern
Intern
Joined: 24 Jun 2019
Posts: 2
Own Kudos [?]: 2 [0]
Given Kudos: 0
Send PM
Re: Any short-cut way to solve this problem? [#permalink]
1
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.
User avatar
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 709 [2]
Given Kudos: 161
Send PM
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.
Verbal Expert
Joined: 18 Apr 2015
Posts: 29960
Own Kudos [?]: 36219 [4]
Given Kudos: 25903
Send PM
Re: If 2^8x=640000 [#permalink]
4
Expert Reply
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

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

\(2^{2x-2} = 5 \sqrt{2}\)
User avatar
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 709 [0]
Given Kudos: 161
Send PM
Re: If 2^8x=640000 [#permalink]
1
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
avatar
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2273 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM
Re: If 2^8x=640000 [#permalink]
1
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 :? )
Manager
Manager
Joined: 04 Jan 2019
Posts: 62
Own Kudos [?]: 114 [0]
Given Kudos: 5
Send PM
Re: If 2^8x=640000 [#permalink]
1
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..
\(2^{8x-10} = 5^{4}\)
Verbal Expert
Joined: 18 Apr 2015
Posts: 29960
Own Kudos [?]: 36219 [0]
Given Kudos: 25903
Send PM
Re: If 2^8x=640000 [#permalink]
Expert Reply
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
Verbal Expert
Joined: 18 Apr 2015
Posts: 29960
Own Kudos [?]: 36219 [0]
Given Kudos: 25903
Send PM
Re: If 2^8x=640000 [#permalink]
Expert Reply
I had been waiting for you Sir and your insight.

Thank you
avatar
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2273 [2]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM
Re: If 2^8x=640000 [#permalink]
2
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 \(2^{2x - 2} = 5 \sqrt{2}\)
Manager
Manager
Joined: 04 Jan 2019
Posts: 62
Own Kudos [?]: 114 [0]
Given Kudos: 5
Send PM
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.
Prep Club for GRE Bot
Re: If 2^8x=640000 [#permalink]
 1   2   
Moderators:
GRE Instructor
78 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne