Question

log2^(5x+3)=log3^(2x+1)

Answer 1

@kc_kennylau

Answer 2

i got it to log2^(5x+3)=log3^(2x+1)

Answer 3

log(a)^b = b*log(a)

Answer 4

so (5x+3)^2=(02x+1)^3
?

Answer 5

Say we have the logarithm with base "b" of an exponential number:
\[\log_{b} a ^{n}\]
by definition we can rewrite the "a^n" as an "a" multiplied "n" times.
\[\log_{b} (a.a.a.a.a.a.a.a...a)\]
there was a property that had this form:
\[(\log_{x} Y)+(\log_{x} Z)=\log_{x} (YZ)\]
applying that to the logarithm:
\[\log_{b} a+\log_{b} a+\log_{b} a+\log_{b} a+\log_{b} a+\log_{b} (a) ... +\log_{b} a\]
but they are all the same factor, for example when we have a+a=2a.
Then, when we have an addition of the same factor, it's the same as the times we sum the factor and multiply it by the variable.
Now, since the Log with base "b" of "a" is added "n" times, then:
\[(n)\log_{b} a\]
Therefore I just proven the following equality:
\[\log_{b} a ^{n} = (n)\log_{b} a\]

Answer 6

applying it to your ptoblem:
\[\log_{} 2^{(5x+3)}=\log_{}3^{(2x+1)} \]

Answer 7

and using the formula I proven:
\[(5x+3)\log_{}2=(2x+1)\log_{} 3\]

Answer 8

that looks very weird

Answer 9

Not really, if we divide both members by log2:
\[\frac{ (5x+3)\log2 }{ \log2 }=\frac{ (2x+1)\log3 }{ \log2 }\]
the "log2" will simplify and if we multiply by the reciprocal "log3" we get:
\[\frac{ 1 }{ (2x+1) }(5x+3)=\frac{ (2x+1)\log3 }{ \log2 }(\frac{ 1 }{ (2x+1) })\]
And if you apply the fractionary multiplication, you will see that (2x+1) on the right side will simplify and you'll end up with:
\[\frac{ 5x+3 }{ 2x+1 }=\frac{ \log3 }{ \log2 }\]

Answer 10

well, it ws not reciprocal "log3" it was reciprocal "2x+1"

Answer 11

Sorry I had to go somewhere but the answer the book gives is \[\frac{ \log (3/8) }{\log (32/9) }\]

Answer 12

your book sucks bro -:(

Answer 13

such a bad form... but, whatever....

Answer 14

it really does tho

Answer 15

\(\log2^{5x+3}=\log3^{2x+1}\)

\((5x + 3)\log2=(2x + 1)\log3\)

\(5x\log2 + 3\log2=2x\log3 + \log3\)

\(5x\log2 - 2x\log3 = \log3 - 3\log2\)

\(x(5\log2 - 2\log3) = \log3 - 3\log2\)

\(x = \dfrac{ \log3 - 3\log2}{5\log2 - 2\log3} \)

Answer 16

that still doesn't give me the answer I'm supposed to get

Answer 17

What answer are you supposed to get?

Answer 18

log2^(5x+3)=log3^(2x+1)
easier solution to this if this is the base of 10
10^(log2^(5x+3))=10^(log3^(2x+1))
2^(5x+3)=3^(2x+1)
solve for x!

Answer 19

Since x is in the exponents on both sides, don't you still need to take log of both sides?

Answer 20

log(3/8)log(32/9)

Answer 21

one over the other scroll up to see it if it doesnt make sense

Answer 22

\(x = \dfrac{ \log3 - 3\log2}{5\log2 - 2\log3}\)

\(x = \dfrac{ \log3 - \log2^3}{\log2^5 - \log3^2}\)

\(x = \dfrac{ \log3 - \log8}{\log32 - \log9}\)

\(x = \dfrac{ \log \frac{3}{8}}{\log \frac{32}{9}}\)

Answer 23

I see. I already continued working on my solution until I got in the form you need.

Answer 24

sounds wierd, but

log _(base 32/9) (3/8)

Answer 25

my bad... this is how you're suppose to do it
log2^(5x+3)=log3^(2x+1)
log2^(5x+3)-log3^(2x+1)=0
split it
5xlog2-2xlog3 +3log2-log3=0
x(5log2-2log3)= log3+3log2

Answer 26

That's exactly what I did. Look above.

Answer 27

yup lol