# How to solve lnx logx 1

## Solve logarithmic equations

Equations that contain logarithms are Logarithmic equations. In the expression lieda(x) are a ≠ 1 and x> 0. Some logarithm equations can be solved using the logarithmic laws. As a rule, an expression that consists of several logarithms must be rewritten in such a way that only one logarithm occurs.

### example 1

Solve the logarithm equation log10(4x + 5) = 2

 log10(4x + 5) = 2 Expose with the base of the logarithm (here: 10) 10log10(4x + 5) = 102 Rewrite with the law of logarithms 4x + 5 = 102 -5 4x = 100 - 5 ÷4 x = \$ \ frac {95} {4} \$ = 23.75 -5

### Example 2

Solve the logarithm equation log10(x + 3) + log10(x) = 1

 log10(x + 3) + log10(x) = 1 Contract logarithms log10(x [x + 3]) = 1 Expose with the base of the logarithm 10log10(x [x + 3]) = 101 Rewrite with the law of logarithms x (x + 3) = 10 Multiply out x2 + 3x = 10 -10 x2 + 3x - 10 = 0 Factoring (x - 2) (x + 5) = 0 To solve => x1 = 2, x2 = -5

Although this equation has two solutions, the logarithm of a negative number is not defined, as is the case with x2 = -5 would be the case. We can check this simply by plugging it back into the original equation. Therefore, this logarithm equation only has the solution x1 = 2

### Example 3

Solve the logarithm equation ln (x - 3) + ln (x - 2) = ln (12x + 24)

 ln (x - 3) + ln (x - 2) = ln (12x + 24) Summarize the sum of two logarithms with the same base as the product of a logarithm ln ((x - 3) (x - 2)) = ln (12x + 24) Bring all logarithms to one page ln ((x - 3) (x - 2)) - ln (12x + 24) = 0 Summarize the difference between two as the quotient of a logarithm \$ \ ln \ left ({\ dfrac {({x-3}) \ cdot ({x-2})} {12x + 24}} \ right) \$ = 0 Expose with the base (here e) and resolve the logarithms \$ {\ dfrac {({x-3}) \ cdot ({x-2})} {12x + 24}} \$ = 1 Multiply both sides by 12x + 24 (x - 3) (x - 2) = 12x + 24 Multiply the left side x2 - 5x + 6 = 12x + 24 Subtract 12x + 24 from both sides x2 - 17x - 18 = 0 Factor the equation (x - 18) (x + 1) = 0 This allows us to split the equation into two new equations x - 18 = 0 or x + 1 = 0 Solve equations x = 18 or x = -1 Replace solutions in the equation and check x = 18

By inserting it again, we find that there is only one solution, namely x = 18.