Question

Solve for x using logarithms. Give an exact answer. Do not enter a decimal approximation.
3times2^x=8times5^x

Answer 1

There are two crucial rules of logarithms needed to solve this, they are as follows:
\[\log(a) + \log(b) = \log(ab)\]\[\log(a^b) = blog(a)\]Okay, using logarithms, you can take a log of each side so that \[3(2^x) = 8(5^x)\]becomes \[\log(3(2^x)) = \log(8(5^x))\]From there using the first rule above we get that \[\log(3) + \log(2^x) = \log(8) + \log(5^x)\] then we see from the second rule that \[\log(3) + xlog(2) = \log(8) + xlog(5)\] Now we just need to tidy it up a little, by moving the x terms to the left, and the constants to the right. \[xlog(2) - xlog(5) = \log(8) - \log(3)\] A touch more confusing, but the first rule can also be applied to subtraction of logarithms, if adding them together results in a multiplication inside the log, then subtracting them gives a division. Like normal algebra you can take the x outside and put brackets so \[x(\log(2/5)) = \log(8/3)\] Now you can just calculate log(2/5) and log(8/3) and treat it as a normal equation to be solved for x :)

Little bonus equation: Most syllabus's don't need you to know this, as it is mostly pointless and can be achieved through other methds but hey, knowledge for knowledge's sake \[\frac{ \log(a) }{ \log(b) } = \log_{b} (a) \] ie. a division resuls in the change of base on the logarithm, but no worries if that's a bit confusing :)