Solving Logs: Wrong Answer vs Calculator Output

[Ry122]

The index laws state that a^m x a^n = a^m+n

These means that in the equation 6^3x-12=6^x
6^3x/6^x = 6^2x
but when I solve for x using logs the answer is different to what my calculator gives me. What am I doing wrong?

 

[VietDao29]

The index laws state that a^m x a^n = a^m+n

These means that in the equation 6^3x-12=6^x
6^3x/6^x = 6^2x
but when I solve for x using logs the answer is different to what my calculator gives me. What am I doing wrong?

Can you please show us what you did? Did you divide both sides by 6^x?

Well, look at it again, it's a cubic equation. :) Can you see it?

Can you go from here? :)

 

[HallsofIvy]

First, those are exponent laws, not "index" laws- indices are just labels distinguishing, say, x₁ from x₂ or x¹ from x². You can't, in general, do arithmetic with indices and there are no "laws" concerning them.

Second, please use parentheses! I guess that when you say a^m x a^n = a^m+n, you really mean a^m x a^n= a^(m+n) because I know that is a law of exponents. But when you write 6^3x-12=6^x I don't know if you mean 6^(3x-12)= 6^x or (6^3x)-12= 6^x!

If you mean (6^3x)- 12= 6^x, then dividing both sides by 6^x doesn't help a lot: you get (6^3x)/6^x- 12/6^x= 6^(2x)- 12(6^{-x})= 1. You could do as VietDao29 suggests: Let y= 6^x so that 6^3x= y^3 and your equation is y^3- 12= y. Similarly, you could divide by 6^x to get 6^(2x)- 12(6^{-x})= 1 and let y= 6^x in that: y^2- 12/y= 1 which is really the sames as y^3- 12= y.

If you mean 6^(3x-12)= 6^x, then you can divide by 6^x to get 6^(2x-12)= 1 implying that 2x-12= 0 (any number to the 0 power equals 1) and x= 6.
Conversely you could take logarithms (base 60 of both sides to get 3x-12= x whence 2x-12= 0 and, again, x= 6.