Solve -4*7^-3.9x=-49

$$- 4 \cdot 7 ^ { - 3.9 x } = - 49$$

Solve for x

$$x=\frac{20\log_{7}\left(2\right)-20}{39}\approx -0.33015016$$

Steps Using Definition of Logarithm

Divide both sides by -4.

$$7^{-3.9x}=\frac{-49}{-4}$$

Fraction \frac{-49}{-4} can be simplified to \frac{49}{4} by removing the negative sign from both the numerator and the denominator.

$$7^{-3.9x}=\frac{49}{4}$$

Take the logarithm of both sides of the equation.

$$\log(7^{-3.9x})=\log(\frac{49}{4})$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$-3.9x\log(7)=\log(\frac{49}{4})$$

Divide both sides by \log(7).

$$-3.9x=\frac{\log(\frac{49}{4})}{\log(7)}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$-3.9x=\log_{7}\left(\frac{49}{4}\right)$$

Divide both sides of the equation by -3.9, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{-2\log_{7}\left(2\right)+2}{-3.9}$$

Solve for x (complex solution)

$$x=-\frac{i\times 20\pi n_{1}}{39\ln(7)}+\frac{20\log_{7}\left(2\right)}{39}-\frac{20}{39}$$

$$n_{1}\in \mathrm{Z}$$