Why do I receive the wrong answer when I try to solve this exponential equation? So I have the equa

hawatajwizp

hawatajwizp

Answered question

2022-06-17

Why do I receive the wrong answer when I try to solve this exponential equation?
So I have the equation: 25 x = 5 x + 6
My reasoning is if you make everything to the base 5:
( 5 2 ) x = 5 x + 5 log 5 6
Given the bases are the same we can do:
2 x = x + log 5 6
x = log 5 6
This answer is wrong however, why is this? Once I've the same bases why can't I do this? Furthermore why couldn't I just take logs of both sides of the original equation?( 25 x = 5 x + 6). What law of logarithms stops me from doing this, why?
Thank you.

Answer & Explanation

Nia Molina

Nia Molina

Beginner2022-06-18Added 21 answers

You can't just 'distribute' the log function over addition. To solve this, notice that
25 x = 5 x + 6
is really just a quadratic equation in disguise
5 2 x 5 x 6 = 0
So let u = 5 x , then we have
u 2 u 6 = ( u 3 ) ( u + 2 ) = 0
Which I'm sure you can easily solve. Once you get the solutions for u, , it is simple to get the solutions for x using 5 x = u.
manierato5h

manierato5h

Beginner2022-06-19Added 5 answers

If y = 5 x , you have y 2 y 6 = ( y 3 ) ( y + 2 ) = 0 so you should be able to solve for y and then x

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