5^x+3 = 25^2x
I know it's something to do with log, but I just can't seem to get it....
2007-12-19 03:38:55 · 10 answers · asked by Tofuie 3 in Science & Mathematics ➔ Mathematics
you take a log of both side so you have
log5^x+3=log25^2x
then you want to get it in the same argument so it then changes too
log5^x+3=log5^4x
then there is a rule to where you can move you exponets on your srguments forward so it becomes
(x+3)log5=(4x)log5
so since they are the same thing except the front then the fronts have to equal
x+3=4x
subtract x from 4x to get 3=3x then divide by 3 to get x by itself and you get
x=1
plug back in you get 5^4=25^2 so 625=625
have a nice day
2007-12-19 03:47:20 · answer #1 · answered by angel_babydoll_devil08 2 · 1⤊ 0⤋
note that 343 = 7^3 and 49 = 7^2 we can rewrite the problem as follows 7^(3(x^2 - 4)) = 7^(2*-3x) since the bases are equal we can now work only with the exponents (since 7^x is injective) 3(x^2 - 4) = 2*-3x 3x^2 - 12 = -6x 3x^2 + 6x - 12 = 0 x^2 + 2x - 4 = 0 x = (-2 +sqrt(4+16))/2 or x = (-2-sqrt(4+16))/2 x = (-2+2sqrt(5))/2 or x = (-2-2sqrt(5))/2 x = sqrt(5) - 1 or x = -1 -sqrt(5) x = 1.23607 or x = -3.23607 note that 36/35 = (6/5)^2 and 125/216 = (5^3/6^3)^-1 = (6/5)^-3 we rewrite the problem as (6/5)^(2(x+3)) = (6/5)^(-3(1/(4-x))) we can equate exponents since bases are equal (and (6/5)^x is injective) we have 2(x+3) = -3/(4-x) 2(x+3)(4-x) = -3 -2x^2 + 2x + 12 = -3 2x^2 - 2x -15 = 0 x = (2+sqrt(4+120))/4 or x = (2-sqrt(4 +120))/4 x = 3.2839 or x = -2.2839
2016-05-25 01:03:06 · answer #2 · answered by kaley 3 · 0⤊ 0⤋
5^x + 3 = 25^2x
5^x + 3 = 5^4x
By taking log
log5^x + 3 = log25^2x
x + 3log5 = 4xlog5
x + 3 = 4x
3x = 3
x = 1
2007-12-19 03:59:02 · answer #3 · answered by Anonymous · 0⤊ 0⤋
5^(x +3) = 25 ^2x
5^(x + 3) = 5 ^ 4x
thus
x+3 = 4x
thus
x=1
2007-12-19 03:44:57 · answer #4 · answered by Senorita 3 · 0⤊ 0⤋
25^2x=5^4x
therefore 4x=x+3
x=1
2007-12-19 03:43:14 · answer #5 · answered by someone else 7 · 0⤊ 0⤋
5^(x+3) = 25^(2x)
Take lg both side
lg 5^(x+3)= lg 25^(2x)
(x+3) lg 5= (2x)lg 25
(x+3)/(2x)= lg 25/lg5
x+3= 2(2x)
x= 1
2007-12-19 03:44:48 · answer #6 · answered by MG 2 · 0⤊ 0⤋
5^(x+3)= 25^(2x)
5^(x+3) = 5^(4x)
x+3 = 4x
3 = 3x
1 = x
2007-12-19 03:44:01 · answer #7 · answered by ironduke8159 7 · 1⤊ 0⤋
log (5^x+3) = log (25^2x)
(x+3)(log5) = (2x)(log25)
Rearrange and get
(x+3) / 2x = (log25 / log5)
(x+3) / 2x = 2
x + 3 = 4x
3x = 3
x = 1
YAY!
2007-12-19 03:51:46 · answer #8 · answered by charly.lizzy 2 · 0⤊ 0⤋
(x+3) ln (5) = 2x ln(25)
(x+3)/2x=ln(25)/ln(5)
(x+3)/2x=ln (5^2)/ln(5)
(x+3)/2x=2 ln(5)/ln(5)
(x+3)/2x=2
(x+3)=4x
x-4x=-3
-3x=-3
x=1
2007-12-19 03:49:55 · answer #9 · answered by cidyah 7 · 0⤊ 0⤋
I am not a calculator
2007-12-19 03:41:29 · answer #10 · answered by Anonymous · 0⤊ 3⤋