Algebra 2 Common Core

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

Chapter 7 - Exponential and Logarithmic Functions - 7-5 Exponential and Logarithmic Equations - Practice and Problem-Solving Exercises - Page 473: 38

Answer

$x=\sqrt{10^5}-1\approx 315.2278$

Work Step by Step

Divide $2$ to both sides: $\dfrac{2\log{(x+1)}}{2}=\dfrac{5}{2}$ $\log{(x+1)}=\frac{5}{2}$ Recall the definition of a logarithm (pg. 451): $\log_{b}{x}=y$ iff $b^y=x$ Applying this definition to our equation (with $b=10, y=5/2, x=x+1$), we get: $10^{\frac{5}{2}}=x+1$ $\left(10^5\right)^{1/2}=x+1$ $\sqrt{10^5}=x+1$ $x=\sqrt{10^5}-1$ $x=10^2\sqrt{10}-1\approx 315.2278$ We confirm that the answer works: $2\log_{10}{(315.2278+1)}=5$ $2\log_{10}{(316.2278)}=5$ $2\cdot 2.5=5$ $5=5$
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