Solve 152x = 36. Round to the nearest ten-thousandth
i am going to make a guess that this is \[15\times 2^x=36\] let me know if that is right or if i am mistaken
yes
ok then first step is to divide by 15 to get \[2^x=\frac{36}{15}=\frac{12}{5}\]
now i think u must change this into logarithmic form :)
then to solve \[v^x=A\iff x=\frac{\ln(A)}{\ln(b)}\] or in your case \[x=\frac{\ln(2.4)}{\ln(2)}\]
typo there, should be \[b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\] in english "the log of the total divided by the log of the base "
since you are going to have to use a calculator it is easier to write \(\frac{12}{5}=2.4\)
\[\LARGE{\log_{2}{12/5}=x}\]
isn't it an easy way @satellite73
@jiteshmeghwal9 it is the same answer, unfortunately you will not get a decimal approximation from this since you do not have log base 2 on your calculator if you want to know the number as decimal you h ave to use the change of base formula for either base e or base 10 because those are the ones available to you
k ! thanx @satellite73 :)
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