Logarithmic progressions

Algebra Level 3

If $1,\log_{9}{(3^{1-x}+2)},\log_{3}{(4.3^x-1)}$ are in Arithmetic progression , then if the value of $x$ is given by $\Large x = \log_{c}{\frac{a}{b}}$ ,then enter your answer as $a+b+c$ .

Note: $.$ $\text{(dot)}$ represents $\times \text{(into)}$

The answer is 10.

1 solution

Kay Xspre
Sep 10, 2015

Arithmetic progression has a fixed constant as common differences. Finding the differences by transforming the equation gives $log_{3} (\frac{\sqrt{3^{1-x}+2}}{3}) = log_{3} (\frac{4(3^{x})-1}{\sqrt{3^{1-x}+2}})$ or more simply $3^{1-x}+5 = 12(3^{x})$ , which equals to $3+5(3^{x})=12(9^{x})$ . Here we can factorize as $(4(3^{x})-3)(3^{x+1}+1) = 0$ Given that $3^{x+1}+1 \neq 0$ , then only $4(3^{x})-3 = 0$ or $x = log_{3}(\frac{3}{4})$ is the answer, hence $a+b+c = 3+3+4 = 10$

A.P has an common difference and not a common ratio.

Sai Ram - 5 years, 9 months ago

Fixed the proper word of "differences" instead of "ratio" (required a lengthy writing as Brilliant will not permit a six-letter "fixed." as the answer)

Kay Xspre - 5 years, 9 months ago

Logarithmic progressions

The answer is 10.

1 solution

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