# Log x 2 + log sqrtx 2 = 9

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

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log x (2) + log sqrtx (2) = 9

We know that :

log a b = log c b /log c a

Let us rewrite log sqrtx (2) using this property.

log sqrtx (2) = log x (2) / log x (sqrtx)

= log x (2) / log x (x)^1/2

= log x (2) / (1/2) log x (x)

= 2 log x (2) / 1

==> log sqrtx (2) = 2*log x (2)

==> log x (2) =2 log x (2) = 9

==> 3*log x (2) = 9

Divide by 3:

==> log x (2) = 3

==> x^3 = 2

==> x= (2)^1/3

william1941 | College Teacher | (Level 3) Valedictorian

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For logarithm we can use the relation log b A= 1/ log A b.

In the question we are given log x 2+ log sqrt x 2 =9

=> 1/ log 2 x+ 1/ log 2 sqrt x =9

=> 1/ log 2 x + 1/ log 2 x^(1/2)= 9

=> 1/ log 2 x+ 1/ (1/2)* log 2 x = 9

=> 1/ log 2 x + 2/ log 2 x= 9

=> 3 / log 2 x= 9

=> 1/ log 2 x = 3

=> log 2 x= 1/3

Therefore x = 2^(1/3)

x= 2^(1/3)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Before solving the equation, we'll create matching bases. We'll switch between the bases and numbers of both logarithms, using the formula:

log a b = 1/log b a

log x 2 = 1/ log 2 x

log sqrtx 2 = 1/log 2 sqrtx

But log 2 sqrt x = log 2 x^1/2

We'll apply the power property to the number log 2 x^1/2.

log 2 x^1/2 = (log 2 x)/2

We'll re-write the equation:

1/log 2 x + 2/log 2 x = 9

We'll substitute log 2 x = t

We'll re-write the equation in t:

1/t + 2/t = 9

1+2=9t

9t = 3

t = 3/9

t = 1/3

log 2 x = 1/3

x = 2^(1/3)

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve logx^2 +logsqrtx^2 = 9.

Solution:

log(x^2) +log (sqrt x^2) = 9

log (x^2) +logx = 9, as sqrt x^2.

log(x^2*x) = 9, as loga+logb = log(ab).

logx^3 = log 10^9.  by denition of log if 10^9 =x, then log x = 9.

Take antilog:

x^3 = 10^9.

Take cube root:

x = {10^9}^(1/3)

x = 10^3 = 1000.