Question

i need precal help!
-1/2+log5/3(4x+3)=1/2

-1+log5/3(2x+1/3)=-3

Answer 1

so can I check, the 1st question is

\[-\frac{1}{2} + \log (\frac{5}{3} (4x + 3) )= \frac{1}{2}\]

Answer 2

what is the base of the log..?

Answer 3

the base is 5/3, it seems odd because ive never seen a log w a fraction as b

Answer 4

@campbell_st

Answer 5

ok... well then its quite easy add 1/2 to both sides and it becomes

\[\log_{\frac{5}{3} }(4x + 3) = 1\]

now you need log laws for powers and bases

\[if ~~~~\log_{a}(x) = b ~~~then~~~~ a^{\log_{a}(x)} = a^b ~~or~~~ x = a^b\]

so by raising both sides of the equation to powers of 5/3

you get

\[\frac{5}{3}^{\log_{\frac{5}{3}}(4x +3)} = (\frac{5}{3})^1\]

which becomes

\[4x + 3 = \frac{5}{3}\]

now solve for x


hope it helps

Answer 6

@campbell_st thank you so much!