expressing logarithms as products - help pls

[sorrel]

so i cant find my notes anywhere on how to do these

express as a product; simplify if possible

log327^-3

the only thing i remember on what to do is move the ^-3 in front of the log
so
-3log327

but i dont know what to do after that

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[Blockzillahead]

yes then

log(3)27 = x

what does 3^? get you 27?

3^3 = 27
log(3)27 = 3

-3(3) = -9

[Ipquarx]

is the ^-3 on the 27 or on the logarithm?

[sorrel]

is the ^-3 on the 27 or on the logarithm?

its on the 27

yes then

awesome! I understand it now, but then after doing the rest I'm stumped on another type of problem.

simplify
log7x^3-log7x

do i put the ^3 in front of the log again?

[BluetoothBoy]

awesome! I understand it now, but then after doing the rest I'm stumped on another type of problem.

simplify
log7x^3-log7x

do i put the ^3 in front of the log again?

Assuming the ^3 is for the log and not the x, yes. It will look like this:

3log₇x - log₇x

[Ipquarx]

And then that simplifies down to either:

2log₇(x) or log₇(x²)

[sorrel]

so then how did you simplify it? from

3log7x - log7x

to this?
2log7(x)

[Ipquarx]

so then how did you simplify it? from

3log7x - log7x

to this?
2log7(x)

You treat them as terms of any regular equation. In other words, 3x - x = 2x, even if x happens to be log₇(x)

[sorrel]

You treat them as terms of any regular equation. In other words, 3x - x = 2x, even if x happens to be log₇(x)

oooh ok thanks, i think i got it now