Question

How do you solve `log_3 (4x-5)=5`?

Answer 1

I got `x=62`

Explanation:

Let us use the definition of log to write:

`4x-5=3^5`

rearrange:

`4x=243+5`

`x=248/4=62`

NB: definition of log
`log_ba=x`
so that:
`a=b^x`

Answer 2

The solution is `x=62`.

Explanation:

Convert the equation to exponential form:

`log_color(red)a(color(green)y)=color(blue)xqquadqquad<=>qquadqquadcolor(red)a^color(blue)x=color(green)y`

Here's the actual equation:

`color(white)=>log_color(red)3(color(green)(4x-5))=color(blue)5`

`=>color(red)3^color(blue)5=color(green)(4x-5)`

`color(white)=>243=4x-5`

`color(white)=>248=4x`

`color(white)=>62=x`

`color(white)=>x=62`

That's the answer. Hope this helped!

Answer 3

Given: `log_3 (4x-5)=5`

Make both sides the exponent of the base, 3:

`3^(log_3 (4x-5))=3^5`

The left side simplifies to the argument of the logarithm and the right side is computed with a calculator:

`4x-5=243`

Solve for x:

`4x = 248`

`x = 62`