1) One of the most important property of logarithmic and exponential functions is that they are inverse of each other and therefore we can convert exponential and logarithmic expressions using the following:
where the symbol \( \iff \) means "is equivalent to", \( y \) is the exponent, \( b \) is the base such that \( b \gt 0 , b \ne 1 \) and \( x \gt 0 \)
Numerical example
\[ 2 = \log_3 (9) \iff 9 = 3^2 \]
2) One-to-one properties of logarithmic and exponential functions
a) If \( \quad b^x = b^y \quad \), then \( \quad x = y \)
b) If \( \quad \log_b (x) = \log_b (y) \quad \), then \( \quad x = y \) , (NOTE: the logs on both sides have the same base)
Questions with Solutions
Convert the given logarithmic expressions into equivalent exponential expressions:
a) \( \log_x (a) = c \)
b) \( \log_b (2x + 1) = 3 \)
Solutions
Convert the given exponential expressions into equivalent logarithmic expressions:
a) \( 3^x = m \)
b) \( x^2 = a \)
Solutions
Evaluate , without calculator, he following logarithmic expressions:
a) \( \log_2 (16 ) \)
b) \( \log_3(27) \)
c) \( \log_2 = \dfrac{1}{32} \)
d) \( \log_{25} 5 \)
e) \( \log \sqrt{10} \)
f) \( \log_b 1 \) , with \( b \gt 0 \) and \( b \ne 0 \)
g) \( \log_{0.1} (10) \)
Solutions
Solve for \( x \) the following logarithmic equations:
a) \( \log_2 x = 3 \)
b) \( \log_x 8 = 3 \)
c) \( \log_3 x = 1 \)
d) \( \log_{5.6} x = 0 \)
e) \( \log_2 (3x + 1) = 4 \)
f) \( \log_3 \dfrac{1}{x+1} = 2 \)
g) \( \log_4 \dfrac{x+1}{2x-1} = 0 \)
h) \( \log ( 1/x + 1 ) = 2 \)
i) \( \log_x 0.0001 = 4 \)
Solutions
Solve for \( x \) the following exponential equations:
a) \( 3^x = 9 \)
b) \( 4^{2x+1} = 16 \)
c) \( \left(\dfrac{1}{2}\right)^x = 2 \)
d) \( 10^x = 5 \)
e) \( \left(\dfrac{1}{3} \right)^{x/2 - 2} = 9 \)
f) \( 0.01^x = 100 \)
g) \( 2^{2x} - 6(2^x) = - 8 \)
Solutions
Solutions to the Above Questions
Solution
Use the equivalent expressions : \( y = log_b(x) \iff x = b^y \) to write
a) \( \log_x (a) = c \) as an exponential \( a = x^c \)
b) \( \log_b (2x + 1) = 3 \) as an exponential \( 2x + 1 = b^3 \)
Solution
Use the equivalent expressions : \( x = b^y \iff y = \log_b (x) \) to write
a) \( 3^x = m \) as a logarithm \( x = \log_3 (m) \)
b) \( x^2 = a \) as a logarithm \( 2 = \log_x (a) \)
Solution
Use the equivalent expressions : \( y = log_b(x) \iff x = b^y \) to evaluate the following without calculator:
a) let \( y = \log_2 16 \)
convert to exponential form: \( 2^y = 16 \)
Use the fact that \( 16 = 2^4\) to write \( 2^y = 2^4 \)
hence using the one to one property of the exponential given above, " If \( 2^y = 2^4 \), then y = 4"
hence : \( y = \log_2 16 = 4 \)
b) let \( y = \log_3 27 \)
convert to exponential form: \( 3^y = 27
Use the fact that \( 27 = 3^3 \) to write \( 3^y = 3^3 \)
hence using the one to one property 2 b) gives then \( y = 3 \)
hence : \( y = \log_3 27 = 3 \)
c) let \( y = \log_2 (1/32) \)
convert to exponential form: \( 2^y = \dfrac{1}{32} \)
Use the fact that \( \dfrac{1}{32} = \dfrac{1}{2^5} = 2^{-5} \) to write \( 2^y = 2^{-5} \)
and the one to one property of the exponential function gives: \( y = -5 \)
hence \( y = \log_2 (1/32) = -5 \)
d) let \( y = \log_{25} 5 \)
convert to exponential form: \( 25^y = 5 \)
Use the fact that \( 5 = \sqrt{25} = 25^{1/2} \) to write \( 25^y = 25^{1/2} \)
hence \( y = \log_{25} 5 = 1/2 \)
e) Let \( y = \log \sqrt{10} \) ( note log base 10 ); convert to exponential form: \( 10^y = \sqrt{10} = 10^{1/2} \), hence \( y = \log \sqrt{10} = 1/2 \)
f) Let \( y = \log_b 1 \) ; convert to exponential form: \( b^y = 1 = b^0 \), hence \( y = log_b 1 = 0 \)
g) let \( y = \log_{0.1} 10 \) ; convert to exponential form: \( 0.1^y = 10 \)
Use \( 10 = 1 / 0.1 = 0.1^{-1} \) , hence \( 0.1^y = 0.1^{-1} \) which gives \( y = \log_{0.1} 10 = - 1 \)
Solution
Use the equivalent expressions : \( y = \log_b(x) \iff x = b^y \) to solve for \( x \) the following logarithmic equations:
a) \( \log_2 x = 3 \) ; convert to exponential form: \( x = 2^3 = 8 \)
b) \( \log_x 8 = 3 \) ; convert to exponential form: \( 8 = x^3 \) , write 8 as \( 8 = 2^3 = x^3\) ; hence x = 2
c)\( \log_3 x = 1 \) ; convert to exponential form: \( x = 3^1 = 3 \)
d) \( \log_{5.6} x = 0 \) ; convert to exponential form: \( x = 5.6^0 = 1 \)
e) \( \log_2 (3x + 1) = 4 \) ; convert to exponential form: \( 3x + 1 = 2^4 = 16 \) , solve for x: \( 3x + 1 = 16 , 3x = 15 , x = 5 \)
f) \( \log_3 \dfrac{1}{x+1} = 2 \) ; convert to exponential form: \( \dfrac{1}{x+1} = 3^2 = 9 \) , solve for x: \( 1 = 9 (x + 1) , x = - 8 / 9 \)
g) \( \log_4 \dfrac{x+1}{2x-1} = 0 \) ; convert to exponential form: \( \dfrac{x+1}{2x-1} = 4^0 = 1 \) , solve for x: \( x + 1 = 2x - 1 , x = 2 \)
h) \( \log ( 1/x + 1 ) = 2 \) ; convert to exponential form: \( 1/x + 1 = 10^2 = 100 \), solve for x: \( 1/x + 1 = 100 , 1/x = 99 , x = 1/99 \)
i) \( \log_x 0.0001 = 4 \) ; convert to exponential form: \( x^4 = 0.0001 = 1/10000 = 1/10^4 = (1/10)^4 \)
which gives \( x^4 = (1/10)^4 \)
hence \( x = 1/10 \).
Solution
Use the one to one property: if \( b^x = b^y \) then \( x = y \) to solve the exponential functions:
Note that in the above equation, the bases of the two exponential are both equal to \( b \).
a) \( 3^x = 9 = 3^2 \) , hence \( x = 2 \)
b) \( 4^{2x + 1} = 16 = 4^2 \), which gives \( 4^{2x + 1} = 4^2 \), hence \( 2x + 1 = 2 , x = 1 / 2 \)
c) \( \left(\dfrac{1}{2} \right)^x = 2 \) , write \( 2 \) as \( 2 = \left(\dfrac{1}{2} \right)^{-1} \), which gives \( \left(\dfrac{1}{2} \right)^x = \left(\dfrac{1}{2} \right)^{-1} \) , hence \( x = - 1 \)
d) \( 10^x = 5 \) , convert the exponential to logarithm base 10: \( \quad x = \log_{10} 5 \) (here we have used convertion to solve the given equation)
e) \( \left(\dfrac{1}{3} \right)^{x/2 - 2} = 9 = 3^2 \). Write \( 3^2 \) as \( \left(\dfrac{1}{3} \right)^{-2} \) , which gives \( \left(\dfrac{1}{3} \right)^{x/2 - 2} = \left(\dfrac{1}{3} \right)^{-2} \). Hence \( x/2 - 2 = - 2 \) , solve to obtain \( x = 0 \)
f) \( 0.01^x = 100 \). Write : \( 100 = \dfrac{1}{0.01} = 0.01^{-1} \) , which gives \( 0.01^x = 0.01^{-1} \), hence \( x = -1 \)
g) \( 2^{2x} - 6 \cdot 2^x = - 8 \)
Note that \( 2^{2x} = (2^x)^2 \).
Let \( u = 2^x \) and write \( u^2 = (2^x)^2 = 2^{2x} \)
We now substitute \( 2^x \) by \( u \) and \( 2^{2x} \) by \( u^2 \) in the given equation and rewrite the equation in terms of \( u \) only and in standard form as follows
\( u^2 - 6 u + 8 = 0 \)
Solve the above quadratic equation by factoring:
\( (u - 2)(u - 4) = 0 \)
solutions in u: \( u = 2 \) and \( u = 4 \)
We now solve for x using the substitution made above:
\( u = 2 = 2^x \) , gives the solution: \( x = 1 \)
\( u = 4 = 2^x = 2^2\) , gives the solution: \( x = 2 \)