1/4 log 2 16 + 1/2 log 2 49 = log 2 x What is the solution to this?
Please help!
@KayJanae I'll help you. Do you know the basic properties of logarithms?
No I dont
okay, the following are important \[1)\log_x y=z \longrightarrow y=x^z\] \[2) a\log b=\log b^a\] \[3) \log x+\log y=\log xy\] \[4) \log x-\log y=\log\frac xy\]
Alright, and do they have a certain name?
Ignore that question lol where do we go from here?
Yeah, let me pull a website, where will you get the names also
Ok, thanks
ok
We have \[\frac 14 \log_2 16 +\frac 1 2 \log_2 49 = \log_2 x\] one more thing, all those properties only if the base is same
Ok, the base is 2 right?
yeah, here all have same, so we could use the properties tell me what's this ? \[\frac 14 \log_2 16\] use one of the properties to simplify
Um, how do i do that?
\[16^{1/4} * 49^{1/2} =x\]
That isnt right. It isnt one of the choices
\[16^{1/4} * 49^{1/2} = 2*7 =14\]
Ohh ok thank you, could you do another one for me?
log 3 0.1 + 2log 3 x = log 3 2 + log 3 5
The principle is the same, just as ash2326 showed you: \[0.1 * x ^{2} = 2 * 5\] that's what your above expression reduces to, remember when you add logs, you multiply the real numbers, and when you subtract logs, you divide the real numbers. can you try solve this one
\[0.1x ^{2} = 10\] \[x ^{2} = 10/0.1 = 100\] \[x=10\]
ok.. i think i understand, so the answer to my question is 10?
yh
Join our real-time social learning platform and learn together with your friends!