How do you find 'x' in this question

Yr 11 Maths

6^(x+1) = 1/36

help please, show me steps and explanation

6 to the power of what gives you 1/36? 6^2=36, and making the exponent negative will give you the reciprocal of that, so it's 6^(-2).

6^(x+1) = 6^(-2)

Take the log(base 6) of both sides. On the left, you should get x+1. On the right, you'll end up with -2.

x + 1 = -2
x = -3

Before you can solve these you need to have the same "base" numbers.

You know that 36 is the same as 6^2 so you can rewrite 1/36 as 1/6^2 OK so far?

but 1/6^2 can also be written as just 6^-2. So now you end up with this equation

6^(x + 1) = 6^-2 now that you have both the bases the same you can simply ignore them and just solve this:

x + 1 = -2

x = -3 DONE.

6 to the power of what gives you 1/36? 6^2=36, and making the exponent negative will give you the reciprocal of that, so it's 6^(-2).
SO,

6^(x + 1) = 6^-2
x + 1 = -2(Take the log(base 6) of both sides. On the left, you should get x+1. On the right, you'll end up with -2.)
x = -3

I may be wrong, but I think you need to do the natural log of both sides

so ln6^x+1 = ln1/36
move x+1 out front
x+1(ln6) = ln1/36
divide by ln6
x+1 = (ln1/36)/(ln6)
subtract by 1
x=(ln1/36)/(ln6) -1
Put this in your calculator and x=-3

Edit: Looks like I did it the hard way. Look at the other guys' method if you don't understand.

6^(x+1)=1/36

Log base 6 both sides.

log (6^(x+1))=log(1/36)

(x+1)log(6)=log(1/36)


x= log(1/36)/log(6)-1=log(6^-2)/log(6^1)-1= -2/1 -1 = -3

6^(x + 1) = 1/36
6^(x + 1) = 1/(6)^2
6^(x + 1) = 6^-2
x + 1 = -2
x = -3