1. ## Solving for n

The numbers log (a^3b^7), log(a^5,b^12) and log(a^8,b^15), are the first three terms of an arithmetic sequence, and the the 12th term of the sequence is log(b^n), find the balue of n.
a) 40
b)56
c)76
d)112
e)143

2. Hello Tiger
Originally Posted by Tiger
The numbers log (a^3b^7), log(a^5,b^12) and log(a^8,b^15), are the first three terms of an arithmetic sequence, and the the 12th term of the sequence is log(b^n), find the balue of n.
a) 40
b)56
c)76
d)112
e)143
The first three terms are:
$\log(a^3b^7),\; \log(a^5b^{12})$ and $\log(a^8b^{15})$
i.e.
$3\log(a) + 7 \log(b),\;5\log(a) + 12\log(b)$ and $8\log(a)+15\log(b)$
The difference between consecutive terms is constant. So:
$2\log(a)+5\log(b) = 3\log(a)+3\log(b)$

$\Rightarrow 2\log(b) = \log(a)$
So the common difference is $2(2\log(b))+5\log(b)=9\log(b)$.

The twelfth term is, therefore:
$3\log(a)+7\log(b) + 11\times 9 \log(b)$
$= 6\log(b) + 7\log(b) + 99\log(b)$

$= 112\log(b)$

$=\log\big(b^{112}\big)$
$\Rightarrow n = 112$

3. Hello, Tiger!

Here's my approach . . .

The numbers $\log(a^3b^7),\;\log(a^5b^{12}),\;\log(a^8b^{15})$
. . are the first three terms of an arithmetic sequence.
The the 12th term of the sequence is $\log(b^n)$

Find the value of $n.$

. . $(a)\;40 \qquad (b)\;56 \qquad (c)\;76 \qquad (d)\;112 \qquad (e)\;143$

We have: . $\begin{Bmatrix}
t_1 &=& \log(a^3b^7) \\ \\[-4mm] t_2 &=& \log(a^5b^{12}) \\ \\[-4mm]

$d \:=\:t_2-t_1 \:=\:\log(a^5b^{12}) - \log(a^3b^7) \:=\: \log\left(\frac{a^5b^{12}}{a^3b^7}\right) \:=\:\log(a^2b^5)\;\;[2]$

$d \:=\:t_3-t_2 \:=\:\log(a^8b^{15}) - \log(a^5b^{12}) \:=\:\log\left(\frac{a^8b^{15}}{a^5b^{12}}\right) \:=\:\log(a^3b^3)\;\;[3]$

Equate [3] and [2]: . $\log(a^3b^3) \:=\:\log(a^2b^5) \quad\Rightarrow\quad a^3b^3 \:=\:a^2b^5 \quad\Rightarrow\quad a \:=\:b^2$

Substitute into [1]: . $\begin{Bmatrix}t_1 &=& \log(b^{13}) \\ t_2 &=& \log(b^{22}) \\ t_3 &=& \log(b^{31}) \end{Bmatrix}$

Substitute into [2]: . $d \:=\:\log(b^9)$

The 12th term is: . $t_{12} \:=\:t_1 + 11d$

We have: . $t_{12} \;\;=\;\;\log(b^{13}) + 11\!\cdot\!\log(b^9) \;\;=\;\;\log(b^{13}) + \log(b^{99}) \;\;=\;\;\log(b^{112})$

Therefore: . $\log(b^n) \:=\:\log(b^{112}) \quad\Rightarrow\quad n \:=\:112$