# What is x if ln x 0

## Solve logarithmic equations

Equations that contain logarithms are **Logarithmic equations**. In the expression lied_{a}(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 log_{10}(4x + 5) = 2

log_{10}(4x + 5) | = | 2 | Expose with the base of the logarithm (here: 10) |

10^{log10(4x + 5)} | = | 10^{2} | Rewrite with the law of logarithms |

4x + 5 | = | 10^{2} | -5 |

4x | = | 100 - 5 | ÷4 |

x | = | $ \ frac {95} {4} $ = 23.75 | -5 |

### Example 2

Solve the logarithm equation log_{10}(x + 3) + log_{10}(x) = 1

log_{10}(x + 3) + log_{10}(x) | = | 1 | Contract logarithms |

log_{10}(x [x + 3]) | = | 1 | Expose with the base of the logarithm |

10^{log10(x [x + 3])} | = | 10^{1} | Rewrite with the law of logarithms |

x (x + 3) | = | 10 | Multiply out |

x^{2} + 3x | = | 10 | -10 |

x^{2} + 3x - 10 | = | 0 | Factoring |

(x - 2) (x + 5) | = | 0 | To solve |

=> | x_{1} = 2, x_{2} = -5 |

Although this equation has two solutions, the logarithm of a negative number is not defined, as is the case with x_{2} = -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 x_{1} = 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 |

x^{2} - 5x + 6 | = | 12x + 24 | Subtract 12x + 24 from both sides |

x^{2} - 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.

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