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logarithmic
- Thread starter Kwesi74
- Start date
Assuming you mean
log<sub>x</sub>(2x+3) = 2
you should have learned that
log<sub>a</sub>(b) = log(b)/log(a) so
log(2x+3)/log(x) = 2
log(2x+3) = 2 log(x) = log(x^2)
If the logs are equal the exponents are also
2x+3=x^2
x^2-2x-3=0
(x+1)(x-3)=0
x=-1 is invalid so x=3
log(4x+1)/log(2) = 3
log(4x+1)= 3*log(2) = log(2^3) log(8)
4x+1=8
x=7/4
log<sub>x</sub>(2x+3) = 2
you should have learned that
log<sub>a</sub>(b) = log(b)/log(a) so
log(2x+3)/log(x) = 2
log(2x+3) = 2 log(x) = log(x^2)
If the logs are equal the exponents are also
2x+3=x^2
x^2-2x-3=0
(x+1)(x-3)=0
x=-1 is invalid so x=3
log(4x+1)/log(2) = 3
log(4x+1)= 3*log(2) = log(2^3) log(8)
4x+1=8
x=7/4
Hello, Kwesi74!
I assume you can rewrite from logs to exponentials: \(\displaystyle \log_b(N)\,=\,N\;\longleftrightarrow\;b^x\,=\,N\)
. . which factors: .\(\displaystyle (x\,-\,3)(x\,+\,1)\:=\:0\)
. . and has roots:. \(\displaystyle x\,=\,3,\,-1\)
But -1 cannot be the base of a logarithm.
Therefore, the only answer is: \(\displaystyle x\,=\,3\)
Then: .\(\displaystyle 4x\:=\:7\;\;\Rightarrow\;\;x\,=\,\frac{7}{4}\)
I assume you can rewrite from logs to exponentials: \(\displaystyle \log_b(N)\,=\,N\;\longleftrightarrow\;b^x\,=\,N\)
Rewrite: .\(\displaystyle x^2\:=\:2x\,+\,3\;\;\Rightarrow\;\;x^2\,-\,2x\,-\,3\:=\:0\)\(\displaystyle \log_x(2x\,+\,3)\:=\:2\)
. . which factors: .\(\displaystyle (x\,-\,3)(x\,+\,1)\:=\:0\)
. . and has roots:. \(\displaystyle x\,=\,3,\,-1\)
But -1 cannot be the base of a logarithm.
Therefore, the only answer is: \(\displaystyle x\,=\,3\)
Rewrite: .\(\displaystyle 2^3\:=\:4x\,+\,1\;\;\Rightarrow\;\;4x\,+\,1\:=\:8\)\(\displaystyle \log_2(4x\,+\,1)\:=\:3\)
Then: .\(\displaystyle 4x\:=\:7\;\;\Rightarrow\;\;x\,=\,\frac{7}{4}\)