Mathematics - Sequence & Series Question with Solution | TestHub

Given that $1$, ${\log }_{9}2(3^x+48)$, and ${\log }_9(3^x-8/3)$ are in an Arithmetic Progression (A.P.).

This implies:

$${\log }_{9}9,\frac{1}{2}{\log }_9(3^x+48),{\log }_9(3^x-8/3)\in A.P.$$

We can rewrite the second term using the logarithm property $n\log a=\log a^n$:

$${\log }_{9}9,{\log }_9(3^x+48{)}^{1/2},{\log }_9(3^x-8/3)\in A.P.$$

Using the property that if $\log a,\log b,\log c\in A.P.$, then $a,b,c\in G.P.$, we can write:

$$9,(3^x+48{)}^{1/2},3^x-\frac{8}{3}\in G.P.$$

For terms in a Geometric Progression (G.P.), the square of the middle term equals the product of the other two terms.

Therefore, we have:

$${\left((3^x+48{)}^{1/2}\right)}^2=9\left(3^x-\frac{8}{3}\right)$$

$$3^x+48=9\left(3^x-\frac{8}{3}\right)$$

Now, let's solve for $x$.

$$3^x+48=9\cdot 3^x-9\cdot \frac{8}{3}$$

$$3^x+48=9\cdot 3^x-24$$

Let $y=3^x$.

$$y+48=9y-24$$

$$48+24=9y-y$$

$$72=8y$$

$$y=\frac{72}{8}$$

$$y=9$$

Substitute back $y=3^x$:

$$3^x=9$$

$$3^x=3^2$$

$$x=2$$