For this case we have the following expression:
![log_ {8} (12) = x-2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/px6sobcjccnw7g1uplakgcwy9d4kx9y4xh.png)
We rewrite how:
![x-2 = log_ {8} (12)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pqdi4kppw6yvvvyrtnhi9ap6z42dx5kcfy.png)
We add 2 to both sides of the equation:
![x = log_ {8} (12) +2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ue1hjn2rnynjqohvwcdfs98w25is2utqas.png)
We rewrite
as
![log_ {8} (2 ^ 2 * 3):](https://img.qammunity.org/2020/formulas/mathematics/middle-school/txn608wxw5ut4ih80wh6z2zf4bfv3bra8c.png)
![x = log_ {8} (2 ^ 2 * 3) +2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/e53l6cfwh5d3aprlv8t8h8gz6ncwj7rkmy.png)
We apply logarithm properties:
![x = log_ {8} (2 ^ 2) + log_ {8} (3) +2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k2vqmv8n2pjrztcs7wr90vovfzrd0fffvb.png)
We apply logarithm properties:
![x = 2log_ {8} (2) + log_ {8} (3) +2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/s320n5oxp6qxjenbi1xbxp88ly30fcfhzv.png)
The logarithmic base 8 of 2 is
![\frac {1} {3}:](https://img.qammunity.org/2020/formulas/mathematics/middle-school/h7mqx13sirli9ump2syrgzy80zx0btrewv.png)
![x = 2 (\frac {1} {3}) + log_ {8} (3) +2\\x = \frac {2} {3} + log_ {8} (3) +2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/w103triqdee6e4frwzxipgozqqwvuifigu.png)
We simplify:
![x = \frac {2} {3} + 2 + log_ {8} (3)\\x = \frac {2 + 6} {3} + log_ {8} (3)\\x = \frac {8} {3} + log_ {8} (3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8uv9mq8b28gk5nysdqgat5954i0phb9k9y.png)
Decimal form:
3,195
Answer:
![x = \frac {8} {3} + log_ {8} (3) = 3,195](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2mh73bj9p0bzdwgcndvvgr0uc4ljbvpr1c.png)