4^(x+3) = 6^(x-2)

[KarlyD]

4^(x+3) = 6^(x-2)

I need help with this! I know I can't get them to the same base, so where do I go from here? I think I need to take the log of the 4 and 6, but I don't know what next.

Please help!

 

[royhaas]

Use the laws of exponents and take the base 10 log of each side.

 

[Denis]

4^(x+3) = 6^(x-2)

rule: a^p * a^q = a^(p+q)

so: 4^x * 4^3 = 6^x * 6^(-2)

rule: a^(-p) = 1 / a^p

so: 4^x * 4^3 = 6^x / 6^2

All yours to wrap up (as per roy)!

 

[KarlyD]

Thank you very much for the help. I'm still having trouble taking the base10 log of each side. (Sorry-I'm really awful at math!) How do I do this??

 

[Deleted member 4993]

Thank you very much for the help. I'm still having trouble taking the base10 log of each side. (Sorry-I'm really awful at math!) How do I do this??

Use a calculator.

4^(x+3) = 6^(x-2)

(x+3) * log(4) = (x-2) * log(6)

(x+3) * 0.602059991 = (x-2) * 0.77815125

Now you can solve for 'x' .....

 

[Denis]

Same idea as Mr K's; just another way (my favorite):

4^(x+3) = 6^(x-2)
4^x * 4^3 = 6^x * 6^(-2)
4^x * 4^3 = 6^x / 6^2
6^x / 4^x = 4^3 * 6^2
(6/4)^x = 64 * 36
(3/2)^x = 2304
x = log(2304) / log(3/2)
x = 19.095112.....