Solve 7+7^2+7^3+7^4+7^5={7}^2x+1

$$7+ { 7 }^{ 2 } + { 7 }^{ 3 } + { 7 }^{ 4 } + { 7 }^{ 5 } = { 7 }^{ 2x+1 }$$

Solve for x

$$x = \frac{\log_{7} {(2801)}}{2} \approx 2.039593601$$

Steps Using Definition of Logarithm

Calculate 7 to the power of 2 and get 49.

$$7+49+7^{3}+7^{4}+7^{5}=7^{2x+1}$$

Add 7 and 49 to get 56.

$$56+7^{3}+7^{4}+7^{5}=7^{2x+1}$$

Calculate 7 to the power of 3 and get 343.

$$56+343+7^{4}+7^{5}=7^{2x+1}$$

Add 56 and 343 to get 399.

$$399+7^{4}+7^{5}=7^{2x+1}$$

Calculate 7 to the power of 4 and get 2401.

$$399+2401+7^{5}=7^{2x+1}$$

Add 399 and 2401 to get 2800.

$$2800+7^{5}=7^{2x+1}$$

Calculate 7 to the power of 5 and get 16807.

$$2800+16807=7^{2x+1}$$

Add 2800 and 16807 to get 19607.

$$19607=7^{2x+1}$$

Swap sides so that all variable terms are on the left hand side.

$$7^{2x+1}=19607$$

Take the logarithm of both sides of the equation.

$$\log(7^{2x+1})=\log(19607)$$

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

$$\left(2x+1\right)\log(7)=\log(19607)$$

Divide both sides by \log(7).

$$2x+1=\frac{\log(19607)}{\log(7)}$$

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

$$2x+1=\log_{7}\left(19607\right)$$

Subtract 1 from both sides of the equation.

$$2x=\log_{7}\left(19607\right)-1$$

Divide both sides by 2.

$$x=\frac{\log_{7}\left(19607\right)-1}{2}$$

Solve for x (complex solution)

$$x=\frac{\pi n_{1}i}{\ln(7)}+\frac{\log_{7}\left(2801\right)}{2}$$

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