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Logarithmic Equations

Logarithmic equations are equations which incorporate logarithm. A logarithm is a function which is the inverse of taking a number to a office. This means that { eq } \log_b a { /eq } is the inverse of { equivalent } b^a { /eq }. Logarithms only take in plus numbers as a value. Given the equation { equivalent } \log_b a = vitamin c { /eq }, this is like saying “ b raised to the power of c equals a. ” Examples of logarithm equations include { equivalent } \log_2 x + 5 = 7 { /eq }, { equivalent } \log_5 5x – \log_5 x – 3 = 1 { /eq }, and { eq } 7\log_3 ( 3x+1 ) + \log_3 ( 3x ) = 14 { /eq }.

The graph of the logarithm basal 2 of adam plus 5 .
The graph of the logarithm base 2 of x plus 5.

Logarithmic Properties

The first key property of the logarithm is that it merely takes in positive values as an input. This means that { eq } \log_ { 10 } 0 { /eq } or { equivalent } \log_ { 10 } ( -1 ) { /eq } are both undefined. In fact, the logarithm function has what is known as a singularity at 0. This means that for any logarithm, as the input approaches 0, the measure of the logarithm approaches negative eternity. never try to take the logarithm of 0 or a negative act. however, the respect of the logarithm after it is evaluated may be a negative number. As mentioned before, the logarithm is the inverse of exponentiation. This means that { eq } \log_b b^a = a { /eq }, and { eq } b^ { \log_b a } = a { /eq }. This property is key in solving some logarithmic equations. These properties are besides used to demonstrate the addition and subtraction properties of logarithm. The total of two logarithm of the lapp establish is { equivalent } \log_b a + \log_b degree centigrade = \log_b ( a * c ) { /eq }. The deviation of two logarithm of the same base is { equivalent } \log_b a – \log_b degree centigrade = \log_b \frac { a } { c } { /eq }. One survive property of logarithm is the multiplication rule : { eq } c\log_b a = \log_b a^c { /eq }.

How to Solve Logarithms

There are several steps involved in solving logarithms. First, group together like terms. This means that numbers outside of the logarithm should be grouped in concert, and that logarithm should be grouped together. then, the properties of logarithm should be applied. This includes using the summation, difference, and generation rules. then, equate both sides and solve for the varying. here is an example of how to solve a logarithm. Consider { equivalent } 2\log_2 ten + \log_2 4 – 3 = 1 { /eq }. First, add 7 to both sides to find that { eq } 2\log_2 ten + \log_2 4= 4 { /eq }. then, apply the generation rule : { eq } \log_2 x^2 + \log_2 4 = 4 { /eq }, and apply the summation rule to find { eq } \log_2 ( 4x^2 ) = 4 { /eq }. Exponentiate both sides to calculate { eq } 4x^2 = 2^4 = 16 { /eq }. then, solve for adam : { eq } x = \sqrt { \frac { 16 } { 4 } } = \sqrt { 4 } = 2 { /eq } .

Solve for X Logarithms: Simple Case

Consider the equation { equivalent } \log_10 ( x+5 ) + 6 = 8 { /eq }. The goal is to solve for x. First, notice that x must be greater than -5. If ten is less than or equal to -5, then the logarithm would be undefined because the logarithm of 0 and negative numbers are undefined.

To solve this equation, foremost subtract the 6 from both sides to find that { eq } \log_10 ( x+5 ) = 2 { /eq }. then, exponentiate both sides to find that { eq } x+5 = 10^2= 100 { /eq }. Hence, it must be that { eq } x = 100 – 5 = 95 { /eq } .

Solve for X Logarithms: Using Product Rule

The product rule, or sum of logarithm, states that { eq } \log_b a + \log_b c = \log_b ( a * c ) { /eq }. This comes from the fact that logarithm are the inverse of exponentiation. Remember the rule for sum of exponents : { eq } b^ { a+c } = b^a * b^c { /eq }. This means that { eq } b^ { \log_b a + \log_b speed of light } = b^ { \log_b a } b^ { \lob_b c } = actinium { /eq }. then, take the logarithm of both sides : { eq } \log_b ( b^ { \log_b a + \log_b carbon } ) = \log_b a + \log_b c = \log_b alternating current { /eq }. The product rule is used when the total of two logarithm appears in a logarithmic equality. Consider the equation { equivalent } \log_2 x + \log_2 2x = 3 { /eq }. then, use the product rule to find that { eq } \log_2 x + \log_2 2x = \log_2 ( x * 2x ) = \log_2 2x^2 = 3 { /eq }. Exponentiate both sides to find that { eq } 2x^2 = 2^3 = 8 { /eq }. Hence, { equivalent } x = \sqrt { \frac { 8 } { 2 } } = \sqrt { 4 } = 2 { /eq }. It is important to notice that although -2 is a solution to { eq } 2x^2 = 8 { /eq }, -2 can not be a solution to the equation because the logarithm of damaging numbers is undefined .

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