Question

Solve the following equation:

`7^2^x-4*7^x=5`

Please explain each step

Answer 1

`x=\log_75`

Explanation:

Given equation:

`7^(2x)-4 \cdot 7^x=5`

`\textsf{Apply exponent rule} \quad a^(bc)=(a^b)^c:`

`\implies (7^x)^2-4 \cdot 7^x=5`

Subtract 5 from both sides:

`\implies (7^x)^2-4 \cdot 7^x-5=0`

`\textsf{Let \; $u = 7^x$}:`

`\implies u^2-4u-5=0`

Factor the quadratic:

`\implies u^2+u-5u-5=0`

`\implies u(u+1)-5(u+1)=0`

`\implies (u-5)(u+1)=0`

Apply the zero-product property:

`\implies u-5=0 \implies u=5`

`\implies u+1=0 \implies u=-1`

`\textsf{Substitute back in \; $u = 7^x$}:`

`\implies 7^x=5`

`\implies 7^x=-1 \quad \textsf{is not valid as $7^x > 0$}.`

Therefore, the only valid solution is:

`\implies 7^x=5`

`\textsf{Apply log law}: \quad a^c=b \iff \log_ab=c`

`\implies \log_75=x`

`\implies x=\log_75`

Answer 2

`x = (\log 5)/(\log 7)`

Explanation:

`7^(2x) - 4 * 7^x = 5`

`7^(2x) - 4 * 7^x - 5 = 0`

`(7^x)^2 - 4(7^x) - 5 = 0`

This is a quadratic equation in 7^x.

Let u = 7^x.

Substitute u for 7^x to see the quadratic equation more clearly.

`u^2 - 4u - 5 = 0`

Now we have a factorable quadratic equation in u.

`(u - 5)(u + 1) = 0`

`u - 5 = 0` or
`u + 1 = 0`

`u = 5` or
`u = -1`

Now substitute back 7^x for u.

`7^x = 5` or
`7^x = -1`

`7^x = -1` does not have a real solution.

Continue solving
`7^x = 5` for x.

Take log of both sides.

`\log 7^x = \log 5`

`x \log 7 = \log 5`

`x = (\log 5)/(\log 7)`