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Solve this equation: log log x = 1The answer is 10^10 but how is it solve?

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islnds | Valedictorian

Posted November 16, 2010 at 7:36 AM via web

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Solve this equation: log log x = 1

The answer is 10^10 but how is it solve?

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

Posted November 16, 2010 at 9:09 AM (Answer #1)

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log log x = 1

Let us assume that :

log x = y

Now substitute into the equation:

log (log x) = 1

==> log y = 1

Now we will rewrite:

==> y = 10^1 = 10

But we know that :

log x = y

==> log x = 10

==> x= 10^10

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giorgiana1976 | College Teacher | Valedictorian

Posted November 16, 2010 at 2:42 PM (Answer #2)

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First, we'll impose constraint of existence of logarithms:

x>0

We know that the inverse function of exponential function is ligarithmic function.

We consider the exponential function:

f(x) = a^x, a>0

a^x = b

To compute x, we'll take decimal logarithms both sides:

log a^x = log b

The power rule of logarithms claims that:

log a^x = x* log a

x* log a = log b

x = log b/log a

If we have the function f(x) = log x

log a x = b

We'll take antilogarithms to calculate x:

x = a^b

So, we'll take antilogarithm to solve the given equation.

log x = 10^1

Log x = 10

We'll take again the antilogarithm, to compute x:

x = 10^10

Remark: Since the base is not indicated, we suppose that the base of both logarithms is 10.

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neela | High School Teacher | Valedictorian

Posted November 16, 2010 at 4:46 PM (Answer #3)

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To solve log logx = 1:

We know if a^b = x, then loga (x) = b.

Alternavely, if loga (x) = b  then x = a^b.

Also remember loga  is normallt used  to write log10 (a), where 10 is the base of the logarithm.

Applying the definition to  log (logx) = 1, we get:   log10 (log10(x) = 1 .

We take thee antilogarithm of log(logx) = 1 and we get:

log10 (x) = 10.

We take the antilogarithm of log10 (x) = 10 and we get:

 x = 10^10.

Now we check whether our result verifies:

Let  x = 10^10.

We take logarithms of both sides:

logx = 10 log 10

logx = 10 , as log10 (10) = 1 by definition.

We take again logarithms of both sides of logx = 10:

log logx = log10.

log (logx) = 1. Or

log logx = 1. 

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