Hi! 4^(x+1) + 3 * 4^x = 112 Should I use log or something?
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Originally Posted by Twig Hi! 4^(x+1) + 3 * 4^x = 112 Should I use log or something? logarithms? no. but good thinking. you should be thinking logarithms for things like this, but in this case, logarithms are not necessary note that can you continue now?
I did solve it, but I doubt I did it "correctly". I got 4 * 4^x + 3 * 4^x = 112 I divided all terms with 4^x so I have 7 = 112/4^x this gives me 4^x = 112/7 now I used logarithms to solve for x
Originally Posted by Twig I did solve it, but I doubt I did it "correctly". I got 4 * 4^x + 3 * 4^x = 112 I divided all terms with 4^x so I have 7 = 112/4^x this gives me 4^x = 112/7 now I used logarithms to solve for x what is in red is completely unnecessary now simplify first. 112/7 = 16 so you have do you really need logarithms for that?
hi so I can basically say that 4 * 4^x + 3* 4^x = 7 * 4^x ? and just from there take 112/7 = 16 and 4^x = 16
Originally Posted by Twig hi so I can basically say that 4 * 4^x + 3* 4^x = 7 * 4^x ? and just from there take 112/7 = 16 and 4^x = 16 yes (you divide both sides by 7) now what is the solution?
x = 2
Originally Posted by Twig x = 2 .....you didn't use logarithms, did you? are we good now? do we understand the concept?
thanks a lot Yes I did. I just didnt realize at first that I can add up my 4^x terms so I get 7 * 4^x
Originally Posted by Twig thanks a lot Yes I did. I just didnt realize at first that I can add up my 4^x terms so I get 7 * 4^x yes. you can essentially treat the 4^x as a variable. just as how if you had 4x + 3x you would get 7x. the idea is the same or if you don't want to think of it that way, you can think of it as factoring out the 4^x
alright cool =)
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