Answer:
![(629)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/de7dm2cgwrpifhr7rbi6n8x27x36ebujzb.png)
Explanation:
So, our equation is:
![log_5((2x-4)/(5)) = 3](https://img.qammunity.org/2023/formulas/mathematics/college/ijjofo2cgd88gvqzhsxtfkyqp2v1137twk.png)
In exponetial form, this looks like:
![5^3=(2x-4)/(5)](https://img.qammunity.org/2023/formulas/mathematics/college/esgl5nkc8jw5izrw9y7g9sttj2uarfq1yb.png)
Now lets cube the 5:
![125=(2x-4)/(5)](https://img.qammunity.org/2023/formulas/mathematics/college/le2ji8doxtgjae7z888f6g3xv47hxp7o90.png)
Next, we can multiply the denominator on the right side by 5:
![625=2x-4](https://img.qammunity.org/2023/formulas/mathematics/college/terli17diqdko1qb8fbpwccwfcspg3xhiv.png)
We need to now get x alone by adding 4 to both sides:
![629=2x](https://img.qammunity.org/2023/formulas/mathematics/college/oua1zq8lwt2q80ixy1c45o177fg9gb93m2.png)
Finally, we divide by x's coefficent, 2, to get:
![x=(629)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/30aany4xti7jmhomi5tu53ag7msz0iw49l.png)
Hope this helps! :3