Logarithm solve

Question

If ( $a_n$) is Geometric sequence and $a_{11}*a_{49} \neq1$ solve
$log_{a_{11}*a_{49}}(a_1*a_2*a_3*...*a_{60})$

My solve:

$(a_{11}*a_{49})^b=(a_1*a_2*a_3*...*a_{60})$

$(a_{1}^2*q^{58})^b=({a_1^2})^{30}*q^{1830}$

What is next?

Answer 1

Since, $a_{n}$ is a G.P. sequence, $$a_{n} = a_{1}r^{n-1}$$

Now, suppose $$b = a_{11}*a_{49}$$

$$\implies b = a_{1}r^{10}*a_{1}r^{48}$$

$$\implies b = a_{1}^{2}r^{58}$$

and, $$a = a_{1}*a_{2}*a_{3}...a_{60}$$

$$\implies a = a_{1}*a_{1}r^{1}*a_{1}r^{2}...*a_{1}r^{59}$$

$$\implies a = a_{1}^{60}*r^{1770}$$

$$\implies a = (a_{1}^{2}r^{59})^{30}$$

$$\implies a = (a_{1}^{2}r^{58}*r)^{30}$$

$$\implies a = (b*r)^{30}$$

then, effectively we have to calculate $$\log_b a$$

$$\implies \log_{b} a = \log_{b} (b*r)^{30}$$

$$\implies \log_{b} a = 30(\log_{b} b + \log_{b} r)$$

$$\implies \log_{b} a = 30(1 + \log_{b} r )$$

Substituting the value of b, answer is $$30*(1 + \log_{a_{11}*a_{49}} r )$$

where, $r$ is the common ratio of the G.P.

Answer 2

since $a_n$ is a geometric series. $a_n = a_0 r^n$

$a_{11}\cdot a_{49} = a_0^2 r^{60}$

$(a_1\cdot a_2\cdots a_{60}) = a_0^{60} r^{\frac 12(60)(61)}$

Here is a rule to help you out $\log_n x = \frac {\log_m x}{\log_m n}$

$\log_{a_{11}\cdot a_{49}} (a_1\cdot a_2\cdots a_{60}) = \frac {\log_r a_0^{60} r^{1830}}{\log_r (a_0^2) r^{60}}\\ \frac {60 \log_r a_0 + 1830}{2\log_r a_0 + 60}$

I actually like the direction you are going...

$(a_{0}^2r^{60})^b=a_0^{60}r^{1830}\\ (a_{0}^2r^{60})^b=(a_0^2r^{61})^{30}\\ (a_{0}^2r^{60})^b=(a_0^2r^{60})^{30}r^{30}\\ b = 30 (1+ \log_{a_{11}\cdot a_{49}} r)$

Answer 3

I think you should write the logarithm in $ln$ terms:

$$\frac{ln(a_1*a_2*...*a_{60})}{ln(a_{11}*a_{49})}$$ then you can use the equations $ln(ab) = ln(a)+ln(b)$ and $ln(q^n) = nln(q)$ you'll get your answer.

$$\frac{ln(a_1*a_2*...*a_{60})}{ln(a_{11}*a_{49})}=\frac{ln(a_1)+...+ln(a_{60})}{ln(a_{11})+ln(a_{49})}=\frac{60ln(a_1)+1830ln(q)}{2ln(a_1)+58ln(q)}$$