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Logarithms Questions

2^(3y-2)=3^(2y+5

7^(2x+5)=7(11^(3x-4)
(How would I get rid of the 7 first?)

3^(2x)=3^(x-1) x 2^(4+x)
Original post by Daydreamer3
1. 2^(3y-2)=3^(2y+5)

2. 7^(2x+5)=7(11^(3x-4)
(How would I get rid of the 7 first?)

3. 3^(2x)=3^(x-1) x 2^(4+x)


Starting with the first one, and as with all of them, split things up!

23y−2=8y42^{3y-2}=\dfrac{8^y}{4} and 32y+5=35⋅9y3^{2y+5} = 3^5\cdot 9^y then divide both sides by either 9y9^y or 8y8^y

Spoiler

(edited 6 years ago)
Original post by RDKGames
Starting with the first one, and as with all of them, split things up!

23y−2=8y42^{3y-2}=\dfrac{8^y}{4} and 32y+5=35⋅9y3^{2y+5} = 3^5\cdot 9^y then divide both sides by either 9y9^y or 8y8^y

Spoiler




Sorry but I dont know what method you're using since we haven't gone that far into logs yet.

I'll show you my method and may you please identify any errors?
Log2 2^(3y-2)=Log2 3^(2y+5)
3y-2=(2y+5)Log2 3
3y-2= 2ylog2 3 + 5log2 3
3y-2ylog2 3 = 2+ 5log2 3
y(3-2log2 3)=2 + 5log2 3
y=5log2 3-2log2 m

(Dont know how to put it into the "maths" form so yeh) By 'Log2 3" I just mean Log small 2 then 3 if that make any sense.
Original post by Daydreamer3
Sorry but I dont know what method you're using since we haven't gone that far into logs yet.

I'll show you my method and may you please identify any errors?
Log2 2^(3y-2)=Log2 3^(2y+5)
3y-2=(2y+5)Log2 3
3y-2= 2ylog2 3 + 5log2 3
3y-2ylog2 3 = 2+ 5log2 3
y(3-2log2 3)=2 + 5log2 3
y=5log2 3-2log2 m

(Dont know how to put it into the "maths" form so yeh) By 'Log2 3" I just mean Log small 2 then 3 if that make any sense.


That' fine, last line should be 2+5log⁡233−2log⁡23\dfrac{2+5\log_2 3}{3-2\log_2 3}
(edited 6 years ago)
Original post by RDKGames
That' fine, last line should be 2+log⁡233−2log⁡23\dfrac{2+\log_2 3}{3-2\log_2 3}


Ohhhh the last line is actually meant to be 2+5log2 3 / 3-2log2 3 whereas I was typing in my calculator what you just wrote as I missed out the 5.

Thank you

Do you know what to do with the 7 in the second question? Do I bring it on the top as a power ? (The 7 in 7(11^(3x-4))
Original post by Daydreamer3
Do you know what to do with the 7 in the second question? Do I bring it on the top as a power ? (The 7 in 7(11^(3x-4))


You can't. Best you can do is divide by it so that the LHS becomes 72x+47^{2x+4}
Original post by RDKGames
You can't. Best you can do is divide by it so that the LHS becomes 72x+47^{2x+4}


Oh okay thank you. So does this only work if the denominator and the numerator are the same number then you can minus 1 from the power?
Original post by Daydreamer3
Oh okay thank you. So does this only work if the denominator and the numerator are the same number then you can minus 1 from the power?


Yes. It should follow from GCSE that anam=an−m\dfrac{a^n}{a^m}=a^{n-m}

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