Question

prefer attachment solve it... pls be fast..​

Answer 1

||✪✪ QUESTION ✪✪||

if 3^(x-1) + 3^(x+1) = 90,, find x ?

|| ✰✰ ANSWER ✰✰ ||

we have ,

3^(x-1) + 3^(x+1) = 90

Solving it by using a^m * a^n = a^(m+n)

→ [ 3^x * 3^(-1) ] + [ 3^x * 3^1 ] = 90

Now , using a^(-b) = 1/a^b

→ [3^x * 1/3 ] + [3^x * 3 ] = 90

Taking 3^x common now, we get,

→ 3^x [ 1/3 + 3 ] = 90

→ 3^x [ (1+9)/3 ] = 90

→ 3^x (10/3) = 90

→ 3^x = 90*3/10

→ 3^x = 27

→ 3^x = 3³

Comparing now, we get, (As , base are same) .

→ x = 3.

Hence, value of X is 3.

Answer 2

Answer:

We know :-

If a/b = c/d From componendo and dividendo ,

we get (a + b) / (a − b )= (c + d) / (c − d)

Applying this property , and get

$$\begin{lgathered}\frac{ {x}^{3} + 3x }{3 {x}^{2} + 1} = \frac{341}{91} \\ = > \frac{ {x}^{3} + 3x + (3 {x}^{2} + 1) }{ {x}^{3} + 3x - (3 {x}^{2} + 1) } = \frac{341 + 91}{341 - 91} \\ = > \frac{ {x}^{3} + 3x + 3 {x}^{2} + 1 }{ {x}^{3} + 3x - 3 {x}^{2} - 1 } = \frac{432}{250} \\ = > \frac{ {x}^{3} + {1}^{3} + 3x(x + 1)}{ {x}^{3} { -1 }^{3} - 3x(x - 1)} = \frac{216}{125} \\ = > \frac{{(x + 1)}^{3} }{ {(x - 1})^{3} } = \frac{ {6}^{3} }{ {5}^{3} } \\ = > \frac{(x + 1)}{(x - 1)} = \frac{6}{5} \\ = > 5x + 5 = 6x - 6 \\ = > 5 + 6 = 6x - 5x \\ = > x = 11\end{lgathered}$$

$$\begin{lgathered}I HOPE ITS HELP YOU DEAR, \\ THANKS\end{lgathered}$$