what is x? 2^(x-1)=log[3]81^2
The answer i got was x=4. I was just wondering if this was correct??
Thanks in advance!
2007-10-25 04:02:04 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Yes
2^(x-1)=log[3] 3^8
2^(x-1)=2^3
x-1=3
x=4
2007-10-25 04:08:36 · answer #1 · answered by iyiogrenci 6 · 2⤊ 0⤋
Yes. You can check by putting 4 into the equation:
2^(4-1)=?log[3]81^2
2^3=?log[3]81^2
3^8=?81^2
3^8=(3^4)^2
3^8=3^8
2007-10-25 04:09:43 · answer #2 · answered by Amelia 6 · 0⤊ 0⤋
If the problem is 2^(x-1)=log[3](81^2), then yes, the answer is 4. If the problem is 2^(x-1)=(log[3]81)^2, then the answer would be 5.
2007-10-25 04:08:55 · answer #3 · answered by chcandles 4 · 0⤊ 0⤋
log(3) 81^2 is 8, so you're looking for an x such that 2^(x-1)=8, so x-1 is 3, and x=4 ... looks fine to me
2007-10-25 04:09:12 · answer #4 · answered by Stephan B 5 · 0⤊ 0⤋
a million) 11x + 2y = - 8 2) 8x + 3y = 5 i could do removing, no longer substitution, to that end Multiply equation a million via 3 and equation 2 via -2 and upload the outcomes 33x + 6y = - 24 -16x - 6y = - 10 - - - - - - - - - - - 17x = - 34 x = - 2 a million) 11(-2) + 2y = - 8 -22 + 2y = - 8 2y = 14 y = 7 verify 8(-2) + 3(7) = 5 -sixteen + 21 = 5 5 = 5
2016-11-09 10:35:51 · answer #5 · answered by ? 4 · 0⤊ 0⤋
simplifying:
3^(2^(x-1))=81^2
make both base equal:
3^(2^(x-1))=(3^4)^2
3^(2^(x-1))=3^8
since bases are the same (3), we can equate their exponents:
2^(x-1) = 8
2^(x-1) = 2^3
x-1 = 3
x=4
2007-10-25 04:32:54 · answer #6 · answered by joyce 2 · 0⤊ 0⤋