logs help

idk how you get y=10^3log+2 according to the mark scheme?

idk how you get y=10^3log+2 according to the mark scheme?

Can you post your attempt?

this is my attempt

How can we "cancel" the logarithm on the LHS, to write the equation in the form $y=f(x)$?

would u write the 2 as log100?

You could, but no point. Just think of an operation that gets rid off the log on the LHS, and apply it on both sides.

i'm not rlly sure tbh, all i can think to do is just to put the 3 up as a power to the x

What's the reverse operation of taking the logarithm??

If I wanted to find $x$ in $2^x=8$ and I would take the logarithm of both sides which gives $\log(2^x)=\log(8)$

So the reverse operation is to...?

take logs of both sides?

No, you'd exponentiate both sides...

If I go back from $\log(2^x)$ I don't take logs to both sides, because then I'd have $\log(\log(2^x))$.

Instead you need to take the expression and put it as the exponent to the base of the logarithm. So $\log(2^x)=\log(8) \Rightarrow 10^{\log(2^x)}=10^{\log(8)}$ and it simplifies from there. (note that $\log$ is the same as $\log_{10}$)

You may wish to review this in your chapter.

yeah but how do i do that with the equation i have currently?
logy=logx^3 +2)

If we exponentiate $x$ (with base 10), then we have $10^x$. How can we therefore exponentiate both sides of this equation?

i rly dont know

$10^{\log_{10}(y)}=y=10^{3\log_{10}(x)+2}$ and simplify RHS.

Again, you may wish to review this chapter - this is a very basic principle.

yeah i'll definitely give it a review

i just don't get how logy turns into that??
apologies for all this