Solve this equation: log log x = 1The answer is 10^10 but how is it solve?
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
First, we'll impose constraint of existence of logarithms:
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.
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.