How do you solve #\log _ { 6} ( 4x - 1) = 4#?

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Answer 1 · 2017-08-07T01:49:19

#x = 1297/4 = 324.25#

We're asked to solve a function for #x#:

#log_6(4x-1) = 4#

We can rearrange this equation into exponential form:

If

#ul(log_(color(red)(b))(color(green)(x)) = color(orange)(a)#

then

#ul(color(red)(b)^(color(orange)(a)) = color(green)(x)#

So

#log_(color(red)(6))(color(green)(4x-1)) = color(orange)(4)#

Is the same as

#color(red)(6)^(color(orange)(4)) = color(green)(4x-1)#

#6^4# is #1296#:

#1296 = 4x-1#

#1297 = 4x#

#color(blue)(ulbar(|stackrel(" ")(" "x = 1297/4 = 324.25" ")|)#