Question

Solve for x.
4^(x+1) = 5^x

Answer 1

variable is in exponent so you can bring it down by taking log both sides

Answer 2

@ganeshie8 So log4^(x+1) = log5^x ?

Answer 3

Yes, but it needs to be a similar base, so log10 but we usually use ln, doesn't really matter. And then the exponent comes down once you take it.\[(x+1)\log_{10}4=xlog_{10}5\]

Answer 4

yes keep going and isolate \(x\)

Answer 5

\[(x+1)\log_{10}4 - x \log_{10} 5 = 0\]

Answer 6

group terms with \(x\)

Answer 7

i would add xlog5 to both sides then divide on both sides to get the x expressions by themselves

Answer 8

\[(x+1)\log ~4 - x \log~ 5 = 0\]

\[x(\log ~4 - \log 5) +\log~ 4 = 0\]

\[x(\log ~4 - \log 5) = -\log~ 4 \]

Answer 9

@ganeshie8 what did you do there? I don't understand how you did that

Answer 10

You'll need to know a few rules to finish it off after ganeshie's work, try solving for x on your own from here, and use this: \[\log_a \left( \frac{M }{ N } \right) =\log_aM-\log_aN\]

Answer 11

He factored the x out after distributing @msasu25

Answer 12

@iambatman ah thank you

Answer 13

:)

Answer 14

@iambatman @ganeshie8 So is the answer, x = -5?

Answer 15

nope you will get a messy expression for x

Answer 16

Yeah, it will involve logs :P

Answer 17

\[(x+1)\log ~4 - x \log~ 5 = 0\]

\[x(\log ~4 - \log 5) +\log~ 4 = 0\]

\[x(\log ~4 - \log 5) = -\log~ 4\]

\[x = -\dfrac{\log~ 4}{\log ~4 - \log 5}\]

\[x = \dfrac{\log~ 4}{\log ~5 - \log 4}\]

\[x = \dfrac{\log~ 4}{\log \left(\frac{5}{4}\right) }\]

Answer 18

you can stop your simplification there ^

Answer 19

Notice, we are never using a specific base of log, so don't bother about it
the base could be any number

Answer 20

@ganeshie8 @iambatman THANK YOU!