Solve the equation. ` log x + log sqrt x + log 2 = 1/4`

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lemjay | High School Teacher | (Level 3) Senior Educator

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`log x + log sqrt x + log 2 = 1/4`

Since the base of the logarithm is not indicated, it indicates that its base is 10.

`log_10x + log_10 sqrtx + log_10 2 = 1/4`

Also, since the logarithms have the same base, express them with one logarithm only. So, apply the product rule `log_b M + log_b N = log_b(MN)` .

`log_10 (x*sqrtx*2) = 1/4`

`log_10(2xsqrtx) = 14`

To simplify, express the square root in exponent form.

`log_10(2x*x^(1/2))=1/4`

To multiply x and x^(1/2), apply the exponent rule `a^m*a^n=a^(m+n)` .

`log_10(2x^(3/2))=1/4`

Then, convert the equation to its exponential form. Take note that the exponential equivalent of `log_b M=a` is `M=b^a` .

`2x^(3/2)=10^(1/4)`

Then, isolate x.

`(2x^(3/2))/2=10^(1/4)/2`

`x^(3/2)=10^(1/4)/2`

`(x^(3/2)^(2/3))=(10^(1/4)/2)^(2/3)`

`x=(10^(1/4)/2)^(2/3)`

To simplify the right side, apply the exponent rules `(a/b)^m=a^m/b^m` and `(a^m)^n=a^(m*n)` .

`x=10^(1/4*2/3)/2^(2/3)`

`x=10^(1/6)/2^(2/3)`

`x=0.925`

Hence, the solution to the given equation is `x=0.0925` . 

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The solution of log x + log sqrt x + log 2 = 1/4 is required.

Use the property of logarithm, log a + log b = log (a*b)

`log x + log sqrt x + log 2 = 1/4`

`log(x*sqrt x*2) = 1/4`

If `log_b x = a` , `x = b^a`

This gives:

`x*sqrt x*2 = 10^(1/4)`

`x^(3/2) = 10^(1/4)/2`

`x = (10^(1/4)/2)^(2/3)`

x = 0.92465

The solution of the equation is approximately x = 0.92465

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