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# Help C2 logarithms watch

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a) Calculate the value of y for which 2 log₃y - log₃(y+4) = 2

b) Calculate the values of z for which log₃z = 4log₃3
2. Use the log laws:

log(a) + log(b) = log(ab)
n*log(a) = log(a^n)

and remember that logs are the opposite/inverse of powers.
3. (Original post by tommm)
Use the log laws:

log(a) + log(b) = log(ab)
n*log(a) = log(a^n)

and remember that logs are the opposite/inverse of powers.
a)2 log₃y - log₃(y+4) = 2
log₃(y^2/y+4)=2

b) log₃z = 4log₃3
log₃z = log₃3^4
log₃z = log₃81
log₃z = 4

Is what I've done correct so far?
4. (Original post by enviusl)
a)2 log₃y - log₃(y+4) = 2
log₃(y^2/y+4)=2

b) log₃z = 4log₃3
log₃z = log₃3^4
log₃z = log₃81
log₃z = 4

Is what I've done correct so far?
Yep, that's all correct

Now, you have something simpler to solve - what happens if you raise both sides to the power 3?
5. Could you not go straight from

to a final solution
6. (Original post by TenOfThem)
Could you not go straight from

to a final solution
Next piece of training. \log looks nicer than log.
7. (Original post by TenOfThem)
See!
8. (Original post by Mr M)
Next piece of training. \log looks nicer than log.
so it does
9. (Original post by enviusl)
a)2 log₃y - log₃(y+4) = 2
log₃(y^2/y+4)=2

b) log₃z = 4log₃3
log₃z = log₃3^4
log₃z = log₃81
log₃z = 4

Is what I've done correct so far?
Yes.

Now you know that if

then

but you could have already seen that after you did line 3 of your working.
10. (Original post by tommm)
Yep, that's all correct

Now, you have something simpler to solve - what happens if you raise both sides to the power 3?
a)2 log₃y - log₃(y+4) = 2
log₃(y^2/y+4)=2
(y^2/y+4) = 3^2
y^2 = 9 (y+4)
y^2 = 9y + 36
y^2 -9y -36 = 0
(y + 3)(y -12)=0
y=-3,12
y=12

b) log₃z = 4log₃3
log₃z = log₃3^4
log₃z = log₃81
log₃z = 4
z=81
11. all correct

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