Introduction:

We will step you how to solving these  equations that have logarithmic expressions.is this logarithmic  equations, that the variable that we are solving for is inside the log expressions.  We will the log equation will rewrite exponentially using the definition of logs to help us get the xoutside of the log.

 

Logarithmic formulas:

 Some properties of logarithms.

 loga(x*y) = loga+ loga

 loga(xy) = y*loga

 loga(x/y) = logax - logay

 logax = logbx / logba

 

Examples:

 

Example:

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Solution:

Example3:

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Solution:

Example4:

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Solution:

Example5:

  into one logarithmic expression

Solution:

 

 

Examples:

 

 into one logarithmic expression

Solution:

Example2:Rewrite  using natural logarithms and evaluate

Solution:

Example3: Express the logarithmic equation  exponentially.

Solution:

Here using formula is 

     if and only if 

.here base is 7    

 we apply the definition we get

Example4: Express the exponential equation   in a logarithmic form.

 Solution:        

   formula is

     if and only if .

Here base is 6

 the exponent is -2,

we get when we put this in log form:

Example5:Evaluate the expression 

Solution:

set the log equal to x

Example6:Evaluate the expression 

Solution:

Example7:Evaluate the expression 

Solution:

Example8: solve log(e^x)=log (e^3)+log (e^5)

Solution

:we know the formula

ln( ex ) = x

Apply the formula

ln( e3 ) = 3 and

 ln( e5 ) = 5

So ln( ex ) = ln( e3 ) + ln( e5 )

x = 3 + 5
x = 8

Example9:Solve the logarithmic equations 3Log7 4+4Log7 3=Log7 X

Solution:

 

3log7(4) + 4log7(3) = log7(x)
 

log7(4^3) + log7(3^4) =log7(x)
 

log7(64) + log7(81) = log7(x)
 

log7(64*81) = log7(x)
 

log7(5184) = log7(x)
 

5184 = x

Example10::Solve the logarithmic equationsLog6 X=1/2Log6 9+1/3Log6 27

Solution:

log6(x) = (1/2)log6(9) + (1/3)log6(27)

log6(x) = log6(9^(1/2)) + log6(27^(1/3))

log6(x) = log6(3) + log6(3)

log6(x) = log6(3*3)

log6(x) = log6(9)

x = 9 

 

Example11:Solve 

Solution:

 use of the properties of logarithms

Now solving equation is 

Now we can change into exponential form is


.

We solve for x

Here we can do factoring 

(x-5)(x+2)=0

x - 5 = 0    or     x + 2 = 0 
x = 5    or     x = -2

Here we take only positive value this is 5

We substitute in original equation

So x=5