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1. Solve the equation: 2log5 (x) - log5 (3x) = 1

am i supposed to factorise the x?
x[2log5(1) - log5 (3)] = 1 ?

or do i have to use the log laws? i'm confused because either way it doesn't seem to work
2. Use log laws to manipulate the LHS so that you have .
Solve the equation: 2log5 (x) - log5 (3x) = 1

am i supposed to factorise the x?
x[2log5(1) - log5 (3)] = 1 ?

or do i have to use the log laws? i'm confused because either way it doesn't seem to work
You can't do that, by factoring out a , you're implying that which is not the case. is a function, whatever goes inside the brackets is called the 'argument' of the function.

To prove to yourself that , try some values for .

Example: Taking , by your definition

Solve the equation: 2log5 (x) - log5 (3x) = 1

am i supposed to factorise the x?
x[2log5(1) - log5 (3)] = 1 ?

or do i have to use the log laws? i'm confused because either way it doesn't seem to work
are you sure it's not 2log5(x)-log5(3x)=-1? I got a solution but it only works for -1

UPDATE: false alarm. I made a mistake. I found the solution. You need to rewrite the 1 as log5(5), move it to the other side of the equation to make it equal 0, and work from there. Feel free to ask in case any more help is needed.
5. (Original post by tim_72)
are you sure it's not 2log5(x)-log5(3x)=-1? I got a solution but it only works for -1
there is a solution to OPs equation, an integer one at that
6. (Original post by DylanJ42)
there is a solution to OPs equation, an integer one at that
7. thank you all guys for the help i solved it!
thank you all guys for the help i solved it!

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