Logarithm Questions - Solve Now!

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To simplify the logarithmic equation, apply the fundamental principle of logarithms, which states that log_a(bc) = log_a(b) + log_a(c). The equation can be rewritten as log_{3a}9 = (log_a9)/(log_a3a) = (log_a3^2)/(log_aa + log_a3) = (2log_a3)/(1 + log_a3). It's important to consistently use LaTeX for mathematical expressions instead of images for clarity. Engaging with the logarithmic principles will enhance understanding and problem-solving skills. Mastering these concepts is crucial for effective algebraic manipulation.
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Homework Statement
By changing the base of log[SUB]3a[/SUB]9, express (log3a9)(1+loga3) as a single logarithm to base a. I don't know what to do to simplify the equation further
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Jouster said:
I don't know what to do to simplify the equation further
Minimize your algebra by applying what you know about logarithms to finish rewriting your first line:
$$\log_{3a}9=\frac{\log_{a}9}{\log_{a}3a}=\frac{\log_{a}3^{2}}{\log_{a}a+\log_{a}3}=\frac{2\log_{a}3}{1+\log_{a}3}$$
 
@Jouster: this is the latest of several threads you've started about logarithms where you have not applied the fundamental principle of logarithms: \log_a(bc) = \log_a(b) + \log_a(c). You need to get into the habit of applying this principle before asking for help.
 
Also @Jouster -- Please post your math using LaTeX, not in images. See the "LaTeX Guide" link below the Edit window for more information. Thank you.
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.