Solve for x (7^(x^2))/(49^x)=343

$\frac{7^{x^{2}}}{49^{x}} = 343$

Step 1

Move $49^{x}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$7^{x^{2}} \cdot 49^{- x} = 343$

Step 2

Rewrite $49$ as $7^{2}$ .

$7^{x^{2}} \cdot {(7^{2})}^{- x} = 343$

Step 3

Multiply the exponents in ${(7^{2})}^{- x}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$7^{x^{2}} \cdot 7^{2 (- x)} = 343$

Multiply $- 1$ by $2$ .

$7^{x^{2}} \cdot 7^{- 2 x} = 343$

Step 4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$7^{x^{2} - 2 x} = 343$

Step 5

Create equivalent expressions in the equation that all have equal bases.

$7^{x^{2} - 2 x} = 7^{3}$

Step 6

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$x^{2} - 2 x = 3$

Step 7

Solve for $x$ .

Subtract $3$ from both sides of the equation.

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

Factor $x^{2} - 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$(x - 3) (x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x + 1 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

Solve for x (7^(x^2))/(49^x)=343

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