Solve the following exponential equation: 62z+3 = 7* using the following methods.(a) Take the "log10" of both sides and simplify. Use your calculator to evaluate logarithms along the way (i.e. get a decimalrepresentation for the logarithm, using 3 decimal places).(b) Take "In" of both sides and simplify. Use your calculator to evaluate logarithms along the way (i.e. get a decimalrepresentation for the logarithm, using 3 decimal places).(c) How close are your answers for (a) and (b)?

Answered: Solve the following exponential…

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

Solve the following exponential equation: 62z+3 = 7* using the following methods.
(a) Take the "log10" of both sides and simplify. Use your calculator to evaluate logarithms along the way (i.e. get a decimal
representation for the logarithm, using 3 decimal places).
(b) Take "In" of both sides and simplify. Use your calculator to evaluate logarithms along the way (i.e. get a decimal
representation for the logarithm, using 3 decimal places).
(c) How close are your answers for (a) and (b)?

Expert Solution

Step 1

Given that

$6^{2 x + 3} = 7^{x}$

(a) Taking " $\log_{10}$ " on both sides we get,

$\log_{10} (6^{2 x + 3}) = \log_{10} (7^{x})$

We will use Logarithm power rule:

$\log_{b} (x^{y}) = y \cdot \log_{b} (x)$

So, we get

$(2 x + 3) \cdot \log_{10} (6) = x \cdot \log_{10} (7)$

Simplifying

$\begin{array}{rcl} \frac{2 x + 3}{x} & = & \frac{\log_{10} (7)}{\log_{10} (6)} \\ 2 + \frac{3}{x} & = & \frac{0.845}{0.778} \\ \frac{3}{x} & = & 1.086 - 2 \\ \frac{3}{x} & = & - 0.914 \\ x & = & \frac{- 3}{0.914} \\ x & = & - 3.282 \end{array}$

Hence we get the value of $x = - 3.282$

Step 2

(b) Taking $" \ln "$ on both sides

$\ln (6^{2 x + 3}) = \ln (7^{x})$

By using logarithm power rule we get

$(2 x + 3) \cdot \ln (6) = x \cdot \ln (7)$

Simplifying

$\begin{array}{rcl} \frac{2 x + 3}{x} & = & \frac{\ln (7)}{\ln (6)} \\ 2 + \frac{3}{x} & = & \frac{1.946}{1.792} \\ \frac{3}{x} & = & 1.086 - 2 \\ \frac{3}{x} & = & - 0.914 \\ x & = & \frac{- 3}{0.914} \\ x & = & - 3.282 \end{array}$