# Solve $x$ $+$ $\log{(1+2^x)}$ $=$ $x\log{5}$ $+$ $\log{6}$

A logarithmic equation is defined in terms of $x$. In this log equation, the sum of variable $x$ and common logarithm of sum of one and two raised to the power of $x$ is equal to the sum of the $x$ times logarithm of five and log of six.

$x$ $+$ $\log{\Big(1+2^x}\Big)$ $\,=\,$ $x\log{5}$ $+$ $\log{6}$

We have to solve this logarithmic equation for evaluating the value of variable $x$.

### Simplify the equation in logarithmic form

The first term in the left hand side of the logarithmic equation is $x$. Similarly, the first term in the right hand side of the equation contains $x$ as a factor. So, we have to bring them to one side for evaluating the variable $x$.

$x$ $+$ $\log{\Big(1+2^x}\Big)$ $\,=\,$ $x\log{5}$ $+$ $\log{6}$

The second terms of both sides of the equation are logarithmic terms. So, they can be brought to another side of the equation.

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

The variable $x$ is a common factor in the both terms in the left hand side expression of the equation. The common factor can be taken out from the both terms of the expression mathematically.

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

Look at the right hand side of the equation. The expression represents the subtraction of two logarithmic terms. It can be simplified by the quotient rule of logarithms.

$\implies$ $x\Big(1-\log{5}\Big)$ $\,=\,$ $\log{\Bigg(\dfrac{6}{1+2^x}}\Bigg)$

The second factor in the left hand side expression is also in subtraction form. It can be simplified, same as the right hand side expression of the equation but it is possible only if we write the number one as the common logarithm of ten.

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

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

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

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

The expression in the left hand side of the equation can be simplified further by the power rule of logarithms.

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

The both sides of the terms are in logarithmic form. Hence, the quantities inside both logarithmic terms should be equal.

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

### Simplify the equation in exponential form

The given logarithmic equation is successfully simplified as an equation in exponential form.

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

Now, let us simplify this exponential equation for finding the variable $x$.

$\implies$ $2^x}(1+2^x}$ $\,=\,$ $6$

Now, use the distributive property for distributing the multiplication over addition in the left hand side of the equation.

$\implies$ $2^x} \times 1+2^x} \times 2^x$ $\,=\,$ $6$

$\implies$ $2^x}+\Big(2^x}\Big)^$ $\,=\,$ $6$

$\implies$ $\Big(2^x}\Big)^2+2^x$ $\,=\,$ $6$

$\implies$ $\Big(2^x}\Big)^2+2^x}-$ $\,=\,$ $0$

Assume that $u = 2^x$ for our convenience. Now, express the equation in terms of $u$ for solving it.

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

The algebraic equation represents a quadratic equation and this quadratic equation 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 $\,u+3 \,=\, 0$

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

We have assumed that $u = 2^x$.

$\,\,\,\therefore\,\,\,\,\,\,$ $2^x} \,=\, 2\$ and $\,2^x} \,=\, -$

### Solve the two exponential equations

We have now two exponential equations and they should be solved for evaluating the $x$. So, let’s first solve the one exponential equation.

$2^x} \,=\,$

In this equation, the bases of both sides of the exponential functions are equal. Hence, their exponents should also be equal.

$\implies$ $2^x} \,=\, 2^$

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

According to the exponential equation $2^x} \,=\,$, it is evaluated that the value of $x$ is equal to $1$.

Now, let us solve the second exponential equation $2^x} \,=\, -$ for finding the value of $x$.

$2^x} \,=\, -$

In this equation, the value $-3$ cannot be expressed in terms of $2$. Hence, we must evaluate the $x$ by logarithms. Now, take the logarithm of both sides of the equation.

$\implies$ $\log{\Big(2^x}\Big) \,=\, \log{(-3)$

$\implies$ $x \times \log{2} \,=\, \log{(-3)}$

Actually, we cannot find the logarithm of a negative number. Hence, the exponential equation $2^x} \,=\, -$ is invalid and the value of $x$ cannot be solved mathematically from this equation.

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

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