What is the Solution for Solving an Equation for y?

[steve snash]

Homework Statement

solve the equation for y,
log5 (14y-12)=2x^6-13

The Attempt at a Solution

ive got no idea how to work it out I am guessing it must be something to do with making the left side with y equal 1, log5 5 =1 but how do i get that?

 

[Mark44]

Every log equation can be written as an equivalent exponential equation, and that's what you want to do here.

The basic idea is that M = log_aN <==> N = a^M

 

[lanedance]

it may help by re-writing the log base b as:
log_b(z) = \frac{ln(z)}{ln(b)}

 

[Mark44]

it may help by re-writing the log base b as:
log_b(z) = \frac{ln(z)}{ln(b)}

Or maybe not. The OP would like to solve for y, which means he needs to get rid of the log part, not write it in some other base.

 

[lanedance]

fair point - i was heading in the same direction, I've just found a lot of people find it easier convert:
X = lnY <==> Y = e^M
rather than
M = log_aN <==> N = a^M
though i know it is really just a redundant step

 

[steve snash]

that makes the equation come up as (14y-12)=5^(2x^6-13)
this then becomes ,
y=(5^(2x^6-13)+12)/14
which i found is the wrong answer =(
is there a way to get rid of the 5^ and simplify it down?

 

[HallsofIvy]

What do you mean "found is the wrong answer"? That is a perfectly good solution to the problem.

 

[steve snash]

really, sweet ty hallsofivy