What is the solution for log x^3 - log10x = log 40?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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Now we are given that log x^3 - log10x = log 40

Use the relation log a - log b = log (a/b)

log x^3 - log10x = log 40

=> log (x^3 /10x) = log 40

=> log x^2/ 10 = log 40

find the anti-log of both the sides

=> x^2/ 10 = 40

=> x^2 = 40*10

=> x^2 = 400

=> x = sqrt (400)

=> x = 20 or x = -20

We don't use x = -20 as the logarithm of a negative number is not defined.

Therefore x = 20

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given the logarithm equation log x^3 - log 10x = log 40.

We need to find x values that satisfies the equation.

We will use logarithm properties to solve.

First, we know that log a - log b = log a/b.

==> log x^3 - log 10x = log 40

==> log x^3/10x = log 40

Now we will simplify.

==> log x^2/10 = log 40

Now, we know that of the logarithms are equal, then the bases are equal too.

Or, if log a = log b ==> a = b

==>  x^2/10 = 40

We will multiply by 10.

==> x^2 = 40*10

==> x^2 = 400

Now we will take the root of both sides.

==> x1= 20

==> x2= -20 ( we will not consider this answer because log 10x is not defined.

Then, the answer is : x = 20

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve for x: logx^3-log10x= log40.

We use  the law of logarithms: loga-logb = log a/b.

Therefore logx^3-log10x= log 40 becomes:

log{x^3/10x} = log(40).

Since loga= logb => a= b, log(x^3/10x)= log40 => x^3/10x = 40.

=> x^2/10 = 40.

=> x^2= 40*10

=> x^2= 400.

=> x = sqrt400.

=> x = 20 , or x= -20. -20 in logarithm is not defined.

 

 

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