How do I solve logarithm equations

log subscript 2 (x-1) = 4

2log subscript 2 (x^2-1) -5 = 1

Can you explain to me how to do it?? Thanks.

To solve log equations, you need to do the following steps:

1. Get the log by itself on one side.

2. Since you have log_2 you need to create a base of 2 on both sides to use the inverse property
2^[log_2(x)] = x

3. Solve for x.

For the first problem, the log is already on one side, so you create a base of 2 on both sides.

2^[log_2(x-1)] = 2^4
x - 1 = 16
x = 17

For the second problem, add 5 to both sides to get:

2log_2(x^2 - 1) = 6

Then, divide by 2 on both sides:

log_2(x^2 - 1) = 3

Create a base of 2 on both sides and solve for x:

2^[log_2(x^2 - 1)] = 2^3
x^2 - 1 = 8
x^2 = 9

So, x = 3 and x = -3 since 3^2 = 9 and (-3)^2 = 9.

1st equation:

def of log subscript a of b = c <=> b = a^c

your case ; a =2, b= x-1, c = 4
hence : x-1 = 2^4
add 1 to equation: x = 16 + 1 = 17


2nd equation:

rule: a log (x) = log(x^a) for whichever subscript

assuming (-5) is not part of the log:

add 5 to equation: 2*log subscript 2 (x^2-1) =6

apply above rule: log subscript 2 (x^2-1)^2 =6

apply def of log: (x^2 - 1 ) ^ 2 = 2^6

rewrite R.H.S.(right hand side) of equal sign in terms of powers of 2: (x^2 - 1)^2 = (2^3)^2

equate bases: x^2 -1 = 2^3 or x^2 - 1 = - 2^3

add 1 to 1st and 2nd equation: x^2 = 8 +1 or x^2 = - 8 + 1

rewrite R.H.S. in terms of powers of 2 (if possible): x^2 = 3 ^ 2 or x^2 = - 7

2nd equation is nonsensical for real numbers => only the 1st is considered

equate bases: x = 3 and - 3 since both 9 = 3^2 and 9 = (-3)^2

log_2 (x - 1) = 4
2^4 = x - 1
16 = x - 1
x = 17

2log_2 (x² - 1) - 5 = 1
2log_2 (x² - 1) = 6
log_2 (x² - 1) = 3
2³ = (x² - 1)
8 + 1 = x²
x² = 9
x = ±3