Exponents

[pkw209]

Hi everyone,

I have a very simple, general question but I was unsure about where to post.

Do you have to have the same bases on both sides of the equation in order to equate their exponents?

i.e. 2^x * 3^4 = 2^6

You can't conclude that x=6 right?

I'm 99.9% sure especially considering the equation 2^6 * 3^4 does not equal 2^6.

Anyway, I'll throw it out there.

Thanks!

[ajith]

Hi everyone,

I have a very simple, general question but I was unsure about where to post.

Do you have to have the same bases on both sides of the equation in order to equate their exponents?

i.e. 2^x * 3^4 = 2^6

You can't conclude that x=6 right?

I'm 99.9% sure especially considering the equation 2^6 * 3^4 does not equal 2^6.

Anyway, I'll throw it out there.

Thanks!

To equate the powers, the bases needs to be same.


2^x * 3^4 = 2^6

2^x = 2^6/3^4 = 64/81

There will be an x , between 0 and -1 which satisfies this equation (2^0 =1 ; 2^-1 is 0.5 ; 64/81 is in between these)

You have to first write 64/81 as a power of 2 to equate powers on both the sides

[shashank.ism]

Hi everyone,

I have a very simple, general question but I was unsure about where to post.

Do you have to have the same bases on both sides of the equation in order to equate their exponents?

i.e. 2^x * 3^4 = 2^6

You can't conclude that x=6 right?

I'm 99.9% sure especially considering the equation 2^6 * 3^4 does not equal 2^6.

Anyway, I'll throw it out there.

Thanks!

Yeah obviously if you get values like 2^x = 2^y
then only u can say that x=y ..otherwise you willl have to solve the way ajith suggested...
But I don't think in exams such difficult situation pops up as of
2^x = 64/81
in that case log vlaue would be surely given in question itself
xlog2 = log (64/81) , so if log value is given then only can solve it

though in real exams only simple eq comes... so just keep in mind if u gonna equate power u will have to keep the bas e same...