Question 1132820: Solve the equation lg(3^x - 2 ^ (4-x )) = 2 + (1/3)lg 8 - (1/4)xlg16
This is how far I got
lg(3^x - 2 ^ (4-x )) = 2 + lg(2^3)^1/3 - lg(2^4)^(1/4)x
= 2 + lg2 - lg2^x
What do I do next? Thanks Answer by Alex.33(110) (Show Source):
You can put this solution on YOUR website! Assume lg refers to the logarithm with a base of 10. I'll show the fastest way I can think up, somewhat similar to where you are at.
lg(3^x-2^(4-x))=2+(1/3)lg8-(1/4)xlg16
lg(3^x-2^(4-x))=lg(100)+(1/3)lg(2^3)-(1/4)xlg(2^4)
lg(3^x-2^(4-x))=lg(100)+lg2-xlg2
lg(3^x-2^(4-x))=lg(200)-lg(2^x)
lg(3^x-2^(4-x))=lg(200/(2^x))
3^x-(2^4/2^x)=200/(2^x)
6^x-16=200
6^x=216
x=3
