Solve 41e^0.3x-27=0

$$41 e ^ { 0.3 x } - 27 = 0$$

Solve for x

$$x=-\frac{10\ln(41)}{3}+10\ln(3)\approx -1.392450669$$

Steps Using Definition of Logarithm

Use the rules of exponents and logarithms to solve the equation.

$$41e^{0.3x}-27=0$$

Add 27 to both sides of the equation.

$$41e^{0.3x}=27$$

Divide both sides by 41.

$$e^{0.3x}=\frac{27}{41}$$

Take the logarithm of both sides of the equation.

$$\log(e^{0.3x})=\log(\frac{27}{41})$$

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

$$0.3x\log(e)=\log(\frac{27}{41})$$

Divide both sides by \log(e).

$$0.3x=\frac{\log(\frac{27}{41})}{\log(e)}$$

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

$$0.3x=\log_{e}\left(\frac{27}{41}\right)$$

Divide both sides of the equation by 0.3, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{\ln(\frac{27}{41})}{0.3}$$

Solve for x (complex solution)

$$x=\frac{i\times 20\pi n_{1}}{3}-\frac{10\ln(41)}{3}+10\ln(3)$$

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