Find X
27^4x-2=243^3x+6
Please explain how to do it.
Thanks
27^(4x-2)=243^(3x+6)
Notice that 27 and 243 are both multiples of 3
Changing them..
(3^4)^(4x-2)=(3^6)^(3x+6)
3^4(4x-2)= 3^6(3x+6)
taking the log base 3 of each side.
(4x-2)= (3x+6)
and solve for x.
That doesn't work if you check it, did you do it wrong?
oops.
27^(4x-2)=243^(3x+6)
Notice that 27 and 243 are both multiples of 3
Changing them..
(3^4)^(4x-2)=(3^6)^(3x+6)
3^4(4x-2)= 3^6(3x+6)
taking the log base 3 of each side.
4(4x-2)= 6(3x+6)
and solve for x.
27^4x-2=243^3x+6 3^3(4x-2)=3^5(3x+6) 3(4x-2)=5(3x+6) 12x-6=15x+30 12x-15x=30+6 -3x=36 x=36/-3 x=-12
To solve the equation, start by distributing the 4 and 6:
16x - 8 = 18x + 36
Next, gather the x terms on one side and constant terms on the other side:
16x - 18x = 36 + 8
Simplifying,
-2x = 44
Divide both sides by -2:
x = -22
So, the value of x in the equation is -22.