log(10,x)<2 — step-by-step solution

1) Let's solve in

ℝ

\log (10, x) < 2

ⓘ log(b, v) is the logarithm of v in base b, that is ln(v)/ln(b):

\log (10, x)

\frac{\ln (x)}{\ln (10)}

\frac{\ln (x)}{\ln (10)} < 2

2) Let's study the sign of:

\ln (10)

\approx 2.3026

\ln (x) < 2 \cdot \ln (10)

ⓘ We subtract

2 \cdot \ln (10)

from both sides

\ln (x) - 2 \cdot \ln (10) < 0

ⓘ For strictly positive arguments, a sum of logarithms is the logarithm of the product, a difference that of the quotient, ln(a) + ln(b) = ln(a*b):

\ln (x) - 2 \cdot \ln (10)

\ln (\frac{x}{10^{2}})

\ln (\frac{x}{10^{2}}) < 0

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

10^{2}

10 \cdot 10

100

\ln (\frac{x}{100}) < 0

Let's apply the exp function on both sides:

e^{\ln (\frac{x}{100})} < e^{0}

ⓘ For real

x

\ln (\exp (x)) = x

; wherever

\ln (x)

is defined,

\exp (\ln (x)) = x

e^{\ln (\frac{x}{100})}

\frac{x}{100}

■Any nonzero number raised to the power 0 is 1:

e^{0}

1

\frac{x}{100} < 1

Candidates, pending the domain check:

x < 100

For the expression

\ln (x)

to be defined, we must have:

x > 0

Domain constraints:

x > 0

▶Solution:

x \in (0, 100)