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Given that $\log_{b}a=0.74$ and $\log_{b}(a-1)=0.65$ find the value of the following expression: $$\log_{b}(a^{4}-1)-2\log_{b}(a^{2}+1)+\log_{b}(a^{3}+a)-\log_{b}(a+1)$$

I tried using log laws to no avail. Help Is much appreciated!

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    $\begingroup$ Start factoring the arguments and use the log laws for cancellation $\endgroup$ – Eleven-Eleven Oct 5 '14 at 6:20
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Using $\log_c(\prod A_i)=\sum\log_c(A_i),$

the given expression can be reduced to $$\log_b(a^2+1)+\log_b(a+1)+\log_b(a-1)-2\log_b(a^2+1)+\log_b(a^2+1)+\log_b(a)-\log_b(a+1)$$

$$=?$$

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Note $$\log_b(a^4-1)-2\log_b(a^2+1)=\log_b((a^2+1)(a^2-1))-\log_b((a^2+1)(a^2+1))$$ $$=\log_b\frac{(a^2+1)(a^2-1)}{(a^2+1)(a^2+1)}$$$$=\log_b(a^2-1)-\log_b(a^2+1)$$ $$=\log_b(a+1)+\log_b(a-1)-\log_b(a^2+1)$$ Now factor the second set of log expressions and simplify using log rules. $$\log_b(a^3+a)-\log_b(a+1)=\log_ba+\log_b(a^2+1)-\log_b(a+1)$$ Finalize by combining and again simplify.
$$\log_b(a+1)+\log_b(a-1)-\log_b(a^2+1)+\log_ba+\log_b(a^2+1)-\log_b(a+1)$$ $$=\log_b(a-1)+\log_ba$$

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  • $\begingroup$ Yeah, I can see what you did with the first half. The second half is weird though. $\endgroup$ – Greg Nelson Oct 5 '14 at 6:52
  • $\begingroup$ I apologize, I had a couple of typos as I did this at 2 in the morning. $\endgroup$ – Eleven-Eleven Oct 5 '14 at 15:44

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