GUYS PLEASE HELP ASAP!? Solve using logarithms. Round the solution to the nearest hundredth. 4^(x)=7^(x+1)
\[\Large 4^x=7^{x+1}\]Taking the natural log of each side gives us,\[\Large \ln\left(4^x\right)=\ln\left(7^{x+1}\right)\]
Our next step will involve this rule of logarithms:\[\Large \color{royalblue}{\log(a^b) \quad=\quad b\cdot \log(a)}\] Do you understand how we can apply this rule?
Yes I'm starting to remember what I learned in that lesson
But can you still show me through it please? I don't wanna make a mistake.
So applying this rule gives us,\[\Large x \ln4\quad=\quad (x+1)\ln7\]See how we applied the rule to bring the exponents outside? :)
The next step is a tad tricky if you don't remember your algebra. We'll want to distribute the ln7 to each term in the brackets.\[\Large x \ln4\quad=\quad x \ln7+\ln7\]
Okay I've got that too :)
Now we want to get all of our x terms on the same side. Let's subtract xln7 from each side.
to continue zepdrix's hint, you get the following: \[x \ln(4) = (x+1)\ln(7) \]then distribute the righ-hand side to get \[x \ln(4) = x \ln(7) + \ln(7)\] then subtract an ln(7) on both sides to get \[x \ln(4) - \ln(7) = x \ln(7)\] now subtract x*ln(4) on both sides to get \[-\ln(7) = x \ln(7) - x \ln(4)\] factor the x from the right side to get \[-\ln(7) = x(\ln(7) - \ln(4))\] to solve for x dived both sides of the equation by \[\ln(7) - \ln(4)\]
lol...
remember...there is a difference between -ln(x) and ln(-x). One is not defined...i'll leave it to you to check.
I don't get the last step you put there honestly :(
Not giving the medal until one person actually can help me out
Oh sorry. Rude guy interrupted. I thought he was going to help you finish this. :p Subtracting xln7 from each side,\[\Large x \ln4-x\ln7\quad=\quad \ln7\]Factoring x out of each term on the left,\[\Large x(\ln4-\ln7)\quad=\quad \ln7\]
The next step is the one you're confused about? We need to isolate the x. So we'll divide each side by the bracket thingy.\[\Large x\quad=\quad \frac{\ln7}{(\ln4-\ln7)}\]
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