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

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Answer 1 · 2018-03-15T00:11:39

I got #x=62#

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 · 2018-03-15T00:14:43

The solution is #x=62#.

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 · 2018-03-15T00:15:16

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#