# Solve $9^x+4$ $\,=\,$ $3^2} \times 27^x+1$

In this problem, it is given that the number $9$ raised to the power of $x$ plus $4$ is equal to $3$ squared times $27$ raised to the power of $x$ plus $1$.

$9^x+4$ $\,=\,$ $3^2} \times 27^x+1$

The value of $x$ has to evaluate by solving the given exponential equation in this exponential equation problem.

### Split each exponential function

There is an exponential function in terms of $x$ on the left-hand side of the equation. Similarly, there is another exponential function in terms of $x$ on the right-hand side of the equation. The two exponential functions are formed by the sum of a variable $x$ and a number. For solving the value of $x$, both exponential functions should be split. It can be done by the product rule of exponents.

$(1).\,\,\,$ $9^x+4$ $\,=\,$ $9^x} \times 9^4$

$(2).\,\,\,$ $27^x+1$ $\,=\,$ $27^x} \times 27^1$

Now, substitute them in the given exponential equation.

$\implies$ $9^x} \times 9^4$ $\,=\,$ $3^2} \times 27^x} \times 27^1$

$\implies$ $9^x} \times 9^4$ $\,=\,$ $3^2} \times 27^x} \times 2$

### d

On the left-hand side of the equation, the bases of both factors are $9$. On the right-hand side of the equation, the base of one factor is $3$ and the bases of the remaining factors are $27$. By factoring, the quantities can be expressed in exponential form with the base of $3$.

$(1).\,\,\,$ $9$ $\,=\,$ $3 \times 3$ $\,=\,$ $3^2$

$(2).\,\,\,$ $27$ $\,=\,$ $3 \times 3 \times 3$ $\,=\,$ $3^3$

Now, substitute them in the exponential equation.

$\implies$ $\Big(3^2}\Big)^x} \times \Big(3^2}\Big)^4$ $\,=\,$ $3^2} \times \Big(3^3}\Big)^x} \times 3^3$

The power of an exponential form quantity can be simplified by the power rule of exponents.

$\implies$ $\Big(3^2 \times x}\Big) \times \Big(3^2 \times 4}\Big$ $\,=\,$ $3^2} \times \Big(3^3 \times x}\Big) \times 3^3$

$\implies$ $3^2x} \times 3^8$ $\,=\,$ $3^2} \times 3^3x} \times 3^3$

### Simplify the exponential equation

$\implies$ $3^2x} \times 3^8$ $\,=\,$ $3^2} \times 3^3} \times 3^3x$

$\implies$ $3^2x+8$ $\,=\,$ $3^2+3+3x$

$\implies$ $3^2x+8$ $\,=\,$ $3^5+3x$

$\implies$ $2x+8$ $\,=\,$ $5+3x$

$\implies$ $5+3x$ $\,=\,$ $2x+8$

$\implies$ $3x-2x$ $\,=\,$ $8-5$

$\,\,\,\therefore\,\,\,\,\,\,$ $x \,=\, 3$

### Shortcut method

$9^x+4$ $\,=\,$ $3^2} \times 27^x+1$

$\Big(3^2}\Big)^x+4$ $\,=\,$ $3^2} \times \Big(3^3}\Big)^x+1$

$3^2 \times (x+4)$ $\,=\,$ $3^2} \times 3^3 \times (x+1)$

$3^2 \times x+2 \times 4$ $\,=\,$ $3^2} \times 3^3 \times x+3 \times 1$

$3^2x+8$ $\,=\,$ $3^2} \times 3^3x+3$

$3^2x+8$ $\,=\,$ $3^2+3x+3$

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