$x$ is a literal number. An equation is formed in terms of logarithm, exponential and algebra. The equation gives us the value of $x$ when the equation is solved.

$x+\log(1+2^x) = x\log 5 +\log 6$

Move $x$ terms to one side and logarithm terms to other side of the equation.

$\implies$ $x-x\log 5 = \log 6 -\log(1+2^{\displaystyle x})$

The literal $x$ is a common in both terms of the left hand side expression. So, take it common from them.

$\implies$ $x(1-\log 5) = \log 6 -\log(1+2^{\displaystyle x})$

The right hand side expression contains two logarithm terms and there is no problem to simplify it but the left hand side expression contain only one logarithm term. So, it is not easy to simplify it. However, there is a technique to do it.

The entire equation is a common logarithm. So, the common logarithm of $10$ is equal to $1$. Hence, write the number $1$ as logarithm of $10$.

$\implies$ $x(\log 10 -\log 5) = \log 6 -\log(1+2^{\displaystyle x})$

The logarithmic terms are in subtraction form in both sides of the equation. The subtraction of two logarithm terms can be simplified by the quotient rule of logarithms.

$\implies$ $x \log \Bigg(\dfrac{10}{5}\Bigg) = \log \Bigg(\dfrac{6}{1+2^{\displaystyle x}}\Bigg)$

$\implies$ $\require{cancel} x \log \Bigg(\dfrac{\cancel{10}}{\cancel{5}}\Bigg) = \log \Bigg(\dfrac{6}{1+2^{\displaystyle x}}\Bigg)$

$\implies$ $x \log 2 = \log \Bigg(\dfrac{6}{1+2^{\displaystyle x}}\Bigg)$

The left hand side expression expresses the product of $x$ and $\log 2$ and it can be transformed by the power rule of logarithms. It is mainly to make both sides of the equation contain logarithmic terms.

$\implies$ $\log {(2^{\displaystyle x})} = \log \Bigg(\dfrac{6}{1+2^{\displaystyle x}}\Bigg)$

Both are logarithmic terms and their bases are also equal. Hence, the numbers in algebraic form of both logarithmic terms are equal mathematically.

$\implies$ $2^{\displaystyle x} = \dfrac{6}{1+2^{\displaystyle x}}$

Now simplify this equation.

$\implies$ $2^{\displaystyle x}(1+2^{\displaystyle x}) = 6$

$\implies$ $2^{\displaystyle x}+ 2^{\displaystyle x} \times 2^{\displaystyle x} = 6$

$\implies$ $2^{\displaystyle x}+ {(2^{\displaystyle x})}^2 = 6$

$\implies$ $2^{\displaystyle x}+ 2^{\displaystyle 2x} = 6$

$\implies$ $2^{\displaystyle 2x} + 2^{\displaystyle x} -6 = 0$

Take $u = 2^{\displaystyle x}$ and transform the entire equation in terms of the literal number $u$.

$\implies u^2 + u -6 = 0$

The algebraic equation represents a quadratic equation and it can be solved in any method but it can be easily solved by the factoring method.

$\implies u^2 +3u -2u -6 = 0$

$\implies u(u+3) -2(u+3) = 0$

$\implies (u-2)(u+3) = 0$

$\implies u-2 = 0$ and $u+3 = 0$

$\therefore \,\,\,\,\,\, u = 2$ and $u = -3$

Therefore, it is derived that there are two values for the literal $u$ and they are $2$ and $-3$ but it is not our required solution for this logarithm problem.

Now, replace the variable $u$ by its real value and it is $2^{\displaystyle x}$.

$\implies 2^{\displaystyle x} = 2 \,\,\,$ and $\,\,\, 2^{\displaystyle x} = -3$

Observe both solutions.

Firstly, $2^{\displaystyle x} = 2$

$\implies 2^{\displaystyle x} = 2^1$

$\therefore \,\,\,\,\,\, x = 1$

The value of exponential term $2^{\displaystyle x}$ can be $2$ if $x$ is equal to $1$.

Now, $2^{\displaystyle x} = -3$

In this equation, $2$ is a positive number and its exponent is unknown. The exponent can be either positive or negative but it represents the number of multiplying terms only. So, the number $2$ can be multiplied to itself any number of times but always gives a positive value and never gives a negative value. Hence, the $2$ raised to the power of $x$ does not equal to $-3$. Therefore, this equation is invalid.

Therefore, the value of $x$ is equal to $1$ and it is the required solution for this logarithm problem.

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