Core Connections Integrated III, 2015
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Core Connections Integrated III, 2015 View details
1. Section 6.1
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Exercise 19 Page 275

Practice makes perfect
a To solve the given logarithmic equation, we will rewrite it as an exponential equation using the definition of a logarithm.
y=log_b x ⇔ x= b^y This definition tells us how to rewrite the logarithm equivalent of y as an exponential equation. The argument x is equal to b raised to the power of y. x=log_(25)( 5) ⇔ 5= 25^x We can see above that x is the exponent to which 25 must be raised to obtain 5. Now, let's solve the exponential equation by rewriting the terms so that they have a common base.
5=25^x
5^1=( 5^2 )^x
5^1=5^(2x)
Now we have two equivalent expressions with the same base. If both sides of the equation are equal, the exponents must also be equal. 5^1=5^(2x) ⇔ 1=2x Finally, we will solve the equation 1=2x.
1=2x
1/2=x
x=1/2
b To solve the given logarithmic equation, we will rewrite it as an exponential equation using the definition of a logarithm.

log_b x=y ⇔ x= b^y This definition tells us how to rewrite the logarithm equivalent of y as an exponential equation. The argument x is equal to b raised to the power of y. log_x( 1)= ⇔ 1= x^() Notice that any x raised to the power of is equal to 1. Since the logarithm base cannot equal 1, x can be any number except for 1.

c To solve the given logarithmic equation, we will rewrite it as an exponential equation using the definition of a logarithm.

y=log_b x ⇔ x= b^y This definition tells us how to rewrite the logarithm equivalent of y as an exponential equation. The argument x is equal to b raised to the power of y. 23=log_(10)( x) ⇔ x= 10^(23)