Write the log expression as a single log. log 2 1 9 x + 2 log 2 3 x \log_2\frac{1}{9x}+2\log_23x lo g 2 9 x 1 + 2 lo g 2 3 x

In this problem, we're asked to write the logarithmic expression as a single log. And here, we're given log base 2 of 1 over 9 x plus 2 times log base 2 of 3 x. So since we want to write this as a single log, that tells us we want to go ahead and condense this. So whenever we're condensing, we always always want to start with our power rule and get anything that is out in front of our logs back into that exponent using that power rule. So the first thing I'm going to do here is bring this 2 back to the exponent of this 3 x. But remember that we need to bring this exponent to the entire thing, not just that x, but the entire 3 x. So this becomes log base 2 of 3x squared, and we move that 2 back into the exponent. So this simplifies to log base 2 of 1 over 9 x because we haven't changed anything there, plus log base 2 of 3x squared, which would give us a 9x squared. Okay. Now that we have applied our power rule, we can go ahead and apply anything else in order to get this down to a single log. Now because I have this addition happening here, that tells me that I'm gonna be using my product rule because I can change addition into multiplication inside of a single log. So let's go ahead and do this. Seeing that these 2 both have the same base, I'm going to have a log base 2 and then multiplying these 2 together, I get 9 x squared over 9 x. Now it looks like I can I can do a little bit more simplifying here? So I can cancel these nines out and then x squared over x, I can get rid of that squared. That x on the bottom will go away and I am simply left with log base 2 of x. And this is my final answer. This 2 log expression has simplified down into just log base 2 of x. Let me know if you have any questions.

Write the log expression as a single log. log ⁡ 2 1 9 x + 2 log ⁡ 2 3 x \log_2\frac{1}{9x}+2\log_23x lo g 2 9 x 1 + 2 lo g 2 3 x

log ⁡ 2 1 3 x \log_2\frac{1}{3x} lo g 2 3 x 1

log ⁡ 2 3 x \log_23x lo g 2 3 x

6. Exponential & Logarithmic Functions

Properties of Logarithms

College Algebra

6. Exponential & Logarithmic Functions

Properties of Logarithms

Struggling with College Algebra?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Write the log expression as a single log.
$\log_2\frac{1}{9x}+2\log_23x$

$\log_2x$

$\log_2\frac{1}{3x}$

$\log_21$

$\log_23x$

Verified step by step guidance

Start by using the property of logarithms that allows you to move the coefficient in front of a logarithm as an exponent inside the logarithm: $ a \cdot \log_b(c) = \log_b(c^a) $. Apply this to $ 2\log_2(3x) $ to get $ \log_2((3x)^2) $.

Next, simplify $ (3x)^2 $ to $ 9x^2 $. So, $ 2\log_2(3x) $ becomes $ \log_2(9x^2) $.

Now, combine the logarithms using the property $ \log_b(a) + \log_b(c) = \log_b(a \cdot c) $. Apply this to $ \log_2\left(\frac{1}{9x}\right) + \log_2(9x^2) $.

Multiply the expressions inside the logarithms: $ \frac{1}{9x} \cdot 9x^2 = x $.

Finally, write the expression as a single logarithm: $ \log_2(x) $.