Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 7 - Exponential and Logarithmic Functions - 7-3 Logarithmic Functions as Inverses - Practice and Problem-Solving Exercises - Page 456: 25



Work Step by Step

Let $x=\log_{49}{7}.$ RECALL: $\log_{a}{y} = b \longleftrightarrow a^b = y.$ Use the definition above to convert the equation to an exponential equation and have $49^x=7.$ Write each side in base 7 to have $(7^2)^x=7 \\7^{2x}=7.$ The bases are the same, so the exponents must be equal. Thus, $2x=1 \\x = \frac{1}{2}.$ Therefore $\log_{49}{7}=\frac{1}{2}$.
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