calling for help (LOGS)
check if its right.
yes its right
the first one is right
the second i have no idea :/
Lets go over it, the first question is \[5^{2x}=10\] \[\log_2 5^{2x} = \log_2 10 \implies x = \frac{ \log_2 10 }{ \log_2 25 }\]
We took the logarithm on both sides so we can bring the x down.
\(Log_{\text{ base }} \text{ Answer} = Exponent\) So \( log_{2}2=1\) So you started off correctly and landed up with the following: \( \log _2( x+3)=4 \) Using the property above we covert the log into an exponent \(2^{4}=x+3\) And then solve for x
wait. So letter a and b are both one problem?
Huh?
we're solving be correct?
No we just did a, we need to do b now.
I explained how to solve part B
okay, I see it now.
so now I have to solve \[_{2}4=x+3\]
YASSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS
NAILED IT!
idk smth doesnt seem right to me
How did you separate \(\log_2(x+3)\) How did you separate the x and the 3 without removing the log?
I first subtracted 1 from both sides Then I simplified 5-1 to 4 and got log2 (x+3)=4
Right you got \[ \log_2 (x+3)=4 \] Now the Log_2 goes on the whole bracket of (x+3) you cant just divide the log_2 to get rid of it. One of the ways to get rid of it is to convert it into an exponent like i showed you above
soso that solution I then have to solve it?
Just figure out what x is?
scroll up and look at my solution
I did.
ok So this is basically how you can convert a log into an exponent \[Log_{ \text{ base}} \text{ Answer } = \text{ exponent }\] Ok here is a general example: \[\log_a b = c \to a^c=b\]
So back to our example: \[\log_2(x+3)=4 \to 2^4=x+3\]
And then we just solve normally \[2^4=x+3\] \[16=x+3\] \[13=x\]
13=x x=13
All the same 13 equals x x equals 13 Both mean the same thing
yea .... I hope you understand it :D
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