If n is an integer and [-3(4n)] = [3(7n -3)], then n=

[leovavassori]

Hello there,

my question relates to the ADVANCED Quant book - Pg 167 -Question #10 In Action Problem Set. It reads:

NOTE: (x) = raised to whatever power is inside the brackets.

10) If n is an integer and [-3(4n)] = [3(7n -3)], then n=

[3(4n)] = [3(7n -3)]
4n = 7n - 3
n= 1

OK so I understand that because n is an integer multiplied by 4 the first 3 on the left side of the equation can only be positive. But what if that was not the case and the first 3 was raised to an odd power, could I still eliminate the bases on both side of equation? Or am I only allowed to do so when the bases are absolutely identical?

Cheers,

Leo

[leovavassori]

Can anyone help please?

[stud.jatt]

No you can't. The reason is that you can only eliminate the base when both sides of the equation have the same base.

In the case where the Left Hand Side of the equation is raised to an odd power the base on the L.H.S is -3 while that on the R.H.S. is 3. And while -3 and 3 might look identical to you, THEY ARE NOT THE SAME NUMBER.

Hello there,

my question relates to the ADVANCED Quant book - Pg 167 -Question #10 In Action Problem Set. It reads:

NOTE: (x) = raised to whatever power is inside the brackets.

10) If n is an integer and [-3(4n)] = [3(7n -3)], then n=

[3(4n)] = [3(7n -3)]
4n = 7n - 3
n= 1

OK so I understand that because n is an integer multiplied by 4 the first 3 on the left side of the equation can only be positive. But what if that was not the case and the first 3 was raised to an odd power, could I still eliminate the bases on both side of equation? Or am I only allowed to do so when the bases are absolutely identical?

Cheers,

Leo

[leovavassori]

Cheers fella. I assumed as much but since I did not come across a case where one of the bases in the equation turned out to be negative I was a bit unsured.

[tim]

cool. let us know if you have any more questions about this one..