0
$\begingroup$

Just a bit confused about how to evaluate the following

$$\log_3 8\times \log_5 9\times \log_2 5$$

What I have done so far:

I have used the change of base rule to change each log to base $3$, so I ended up with this after cancelling: $$\log_3 8\cdot \log_3 9 \cdot \frac{1}{\log_32}$$

First of all, I'm not even sure that I have made the right decision in changing to base 3, does it matter what you change to? Second of all, I'm just unsure where to go from here.

$\endgroup$
1
  • 1
    $\begingroup$ It doesn't matter which base you choose. Now you can write that $8=2^3$ and $9=3^2$ and use the rule $\log_a \left(b^c\right)=c\log_a b$. Good luck! $\endgroup$
    – Galc127
    May 25, 2016 at 10:28

2 Answers 2

2
$\begingroup$

To systematically attack such questions, use $\log_x y = \frac{\log y}{ \log x}$ i.e. $$\log_3 8 = \frac{\ln 8}{\ln 3}, \quad \log_5 9 = \frac{\ln 9}{\ln 5}, \quad \log_2 5 = \frac{\ln 5}{\ln 2}$$ multiply and simplify $$\log_3 8 \times \log_5 9 \times \log_2 5 = \frac{\ln 8}{\ln 3} \times \frac{\ln 9}{\ln 5} \times \frac{\ln 5}{\ln 2}\\ =\frac{\ln 8 \times \ln 9}{\ln 3 \times \ln 2} =\frac{\ln 8}{\ln 2} \times \frac{\ln 9}{\ln 3} = 3\times2 = 6$$

$\endgroup$
0
$\begingroup$

Notice:

$$\log_a(b)=\frac{\ln(b)}{\ln(a)}$$ $$\ln(a^b)=b\ln(a)$$

So:

$$\log_3(8)\cdot\log_5(9)\cdot\log_2(5)=\frac{\ln(8)}{\ln(3)}\cdot\frac{\ln(9)}{\ln(5)}\cdot\frac{\ln(5)}{\ln(2)}=\frac{\ln(8)}{\ln(3)}\cdot\frac{\ln(9)}{1}\cdot\frac{1}{\ln(2)}=\frac{\ln(8)}{\ln(3)}\cdot\frac{\ln(9)}{\ln(2)}=$$ $$\frac{\ln(8)}{\ln(3)}\cdot\frac{2\ln(3)}{\ln(2)}=\frac{\ln(8)}{1}\cdot\frac{2}{\ln(2)}=\frac{3\ln(2)}{1}\cdot\frac{2}{\ln(2)}=\frac{3\ln(2)}{\ln(2)}=3\cdot2=6$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .