# log x^2 = 15 + log 10

Asked on by corrcorina

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

Posted on

log x^2 = 15 + log 10

We know that log 10 = 1

==> log x^2 = 15 + 1

==> log x^2 = 16

==> x^2 = 10^16

==> x = sqrt10^16

= sqrt(10^8)^2

= 10^8

==> x= 10^8

Top Answer

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We could write 15 = 15*1 = 15*log 10

log x^2 = log 10 + 15*log 10

We'll factorize:

log x^2 = log 10*(1+15)

log x^2 = 16*log 10

We'll use the power property of logarithms:

log x^2 = log 10^16

We'll use the one to one property and we'll get:

x^2 = 10^16

x1 = +sqrt 10^16

x1 = +10^8

x2 = -10^8

For log x^2 to exist, x>0, so the equation will have only one solution, namely x = +10^8

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve logx^2 = 15 +log10.

Solution:

logx^2 = 15+log10

logx^2 = log10^15 +log10

logx^2 = log (10^15 * 10), as loga+logb = logab. Take antilog.

x^2 = 10^15 *10 = 10^16.

x = sqrt10^16 = 10^8

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