Question

find 3^x by solving the expression
$log_{5 }((3 {}^{2x} +4 ) \div (3 {}^{x} + 2)) = 1$
​

Answer 1

Answer:2

Step-by-step explanation:

Answer 2

The value of $3^{x}$ is $2$.

Concept to be used:

Before we solve this problem, we must be well aware of some concepts that we are going to use.

Exponents:

$(a^{m})^{n}=a^{m\times n}=a^{mn}$

Logarithmic:

We cannot find log of any negative value. So, $log(-a)$ does not exist, where $a>0$.

Also,

• $log(a^{m})=m\:log(a)$

• $\dfrac{log(a)}{log(b)}=log_{b}(a)$

Step-by-step explanation:

Here, $(3^{2x}+4)\div (3^{x}+2)=1$

$\Rightarrow \dfrac{3^{2x}+4}{3^{x}+2}=1$

$\Rightarrow 3^{2x}+4=3^{x}+2$

• Hint. We have transposed $(3^{x}+2)$ to the right hand side.

$\Rightarrow 3^{2x}-3^{x}-2=0$

• Hint. We have transposed all the terms to the left hand side.

$\Rightarrow (3^{x})^{2}-3^{x}-2=0$

• Hint. We intend to factorize the terms.

$\Rightarrow (3^{x})^{2}-2\times 3^{x}+3^{x}-2=0$

$\Rightarrow 3^{x}(3^{x}-2)+1(3^{x}-2)=0$

$\Rightarrow (3^{x}-2)(3^{x}+1)=0$

Either $3^{x}-2=0$ or, $3^{x}+1=0$

$\Rightarrow 3^{x}=2$ or, $3^{x}=-1$

Further:

Since logarithm of a negative number does not exist, taking $log$ for the first equation, we have

$log(3^{x})=log(2)$

$\Rightarrow x\:log(3)=log(2)$

• Hint. Since $log(a^{m})=m\:log(a)$

$\Rightarrow x=\dfrac{log(2)}{log(3)}$

$\Rightarrow x=log_{3}(2)$

Conclusion:

From 'Further' part, we can say that only $3^{x}=2$ is possible.

#SPJ3