Solve for x 3^(1-2x)=4^x

$3^{1 - 2 x} = 4^{x}$

Step 1

Take the log of both sides of the equation.

$\ln (3^{1 - 2 x}) = \ln (4^{x})$

Step 2

Expand $\ln (3^{1 - 2 x})$ by moving $1 - 2 x$ outside the logarithm.

$(1 - 2 x) \ln (3) = \ln (4^{x})$

Step 3

Expand $\ln (4^{x})$ by moving $x$ outside the logarithm.

$(1 - 2 x) \ln (3) = x \ln (4)$

Step 4

Solve the equation for $x$ .

Simplify $(1 - 2 x) \ln (3)$ .

Rewrite.

$0 + 0 + (1 - 2 x) \ln (3) = x \ln (4)$

Simplify by adding zeros.

$(1 - 2 x) \ln (3) = x \ln (4)$

Apply the distributive property.

$1 \ln (3) - 2 x \ln (3) = x \ln (4)$

Multiply $\ln (3)$ by $1$ .

$\ln (3) - 2 x \ln (3) = x \ln (4)$

Subtract $x \ln (4)$ from both sides of the equation.

$\ln (3) - 2 x \ln (3) - x \ln (4) = 0$

Subtract $\ln (3)$ from both sides of the equation.

$- 2 x \ln (3) - x \ln (4) = - \ln (3)$

Factor $x$ out of $- 2 x \ln (3) - x \ln (4)$ .

Factor $x$ out of $- 2 x \ln (3)$ .

$x (- 2 \ln (3)) - x \ln (4) = - \ln (3)$

Factor $x$ out of $- x \ln (4)$ .

$x (- 2 \ln (3)) + x (- 1 \ln (4)) = - \ln (3)$

Factor $x$ out of $x (- 2 \ln (3)) + x (- 1 \ln (4))$ .

$x (- 2 \ln (3) - 1 \ln (4)) = - \ln (3)$

Rewrite $- 1 \ln (4)$ as $- \ln (4)$ .

$x (- 2 \ln (3) - \ln (4)) = - \ln (3)$

Divide each term in $x (- 2 \ln (3) - \ln (4)) = - \ln (3)$ by $- 2 \ln (3) - \ln (4)$ and simplify.

Divide each term in $x (- 2 \ln (3) - \ln (4)) = - \ln (3)$ by $- 2 \ln (3) - \ln (4)$ .

$\frac{x (- 2 \ln (3) - \ln (4))}{- 2 \ln (3) - \ln (4)} = \frac{- \ln (3)}{- 2 \ln (3) - \ln (4)}$

Simplify the left side.

Cancel the common factor of $- 2 \ln (3) - \ln (4)$ .

Cancel the common factor.

$\frac{x (- 2 \ln (3) - \ln (4))}{- 2 \ln (3) - \ln (4)} = \frac{- \ln (3)}{- 2 \ln (3) - \ln (4)}$

Divide $x$ by $1$ .

$x = \frac{- \ln (3)}{- 2 \ln (3) - \ln (4)}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{\ln (3)}{- 2 \ln (3) - \ln (4)}$

Factor $- 1$ out of $- 2 \ln (3)$ .

$x = - \frac{\ln (3)}{- (2 \ln (3)) - \ln (4)}$

Factor $- 1$ out of $- \ln (4)$ .

$x = - \frac{\ln (3)}{- (2 \ln (3)) - \ln (4)}$

Factor $- 1$ out of $- (2 \ln (3)) - \ln (4)$ .

$x = - \frac{\ln (3)}{- (2 \ln (3) + \ln (4))}$

Simplify the expression.

Rewrite $- (2 \ln (3) + \ln (4))$ as $- 1 (2 \ln (3) + \ln (4))$ .

$x = - \frac{\ln (3)}{- 1 (2 \ln (3) + \ln (4))}$

Move the negative in front of the fraction.

$x = - - \frac{\ln (3)}{2 \ln (3) + \ln (4)}$

Multiply $- 1$ by $- 1$ .

$x = 1 \frac{\ln (3)}{2 \ln (3) + \ln (4)}$

Multiply $\frac{\ln (3)}{2 \ln (3) + \ln (4)}$ by $1$ .

$x = \frac{\ln (3)}{2 \ln (3) + \ln (4)}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (3)}{2 \ln (3) + \ln (4)}$

Decimal Form:

$x = 0.30657359 \dots$

Solve for x 3^(1-2x)=4^x

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