I need help trying to figure out how to solve
log (2-3x) + log (3-2x) = 1.5
logbase7(1-x) - logbase7(x+2) = logbase7(x^2)
Thank You!
Hello,
You know that log(a)+log(b)=log(ab)log (2-3x) + log (3-2x) = 1.5
So $\displaystyle \log(2-3x)+\log(3-2x)=\log((2-3x)(3-2x))=1.5$
And you can continue by developping what's in brackets and taking the exponential (beware of the domain of the function)
Since $\displaystyle \log_{x} y = \frac{\log(y)}{\log{x}}$, you can transform the previous equation :logbase7(1-x) - logbase7(x+2) = logbase7(x^2)
$\displaystyle \log_7 (1-x) - \log_7 (x+2) = log_7 (x^2) \Longleftrightarrow \log (1-x) - \log(x+2) = \log(x^2)$
Since log(a)-log(b)=log(a/b), it's the same as :
$\displaystyle \log(\frac{1-x}{x+2}) = \log(x^2)$
And if log(a)=log(b), then a=b