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 $\displaystyle 4^{x + 1} = 4 \cdot 4^x$ 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 $\displaystyle 4^x = 16$ 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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