Let's solve each equation and express the answers to three decimal places.

a) $6^x = 78$

Take the natural logarithm (ln) of both sides: $\ln(6^x) = \ln(78)$

Using the logarithmic property $\ln(a^b) = b \ln(a)$: $x \ln(6) = \ln(78)$

Solve for $x$: $x = \frac{\ln(78)}{\ln(6)}$

b) $(5.4)^x = 234$

Take the natural logarithm of both sides: $\ln(5.4^x) = \ln(234)$

Using $\ln(a^b) = b \ln(a)$: $x \ln(5.4) = \ln(234)$

Solve for $x$: $x = \frac{\ln(234)}{\ln(5.4)}$

c) $8(3^x) = 132$

Divide through by 8 to isolate $3^x$: $3^x = \frac{132}{8} = 16.5$

Take the natural logarithm of both sides: $\ln(3^x) = \ln(16.5)$

Using $\ln(a^b) = b \ln(a)$: $x \ln(3) = \ln(16.5)$

Solve for $x$: $x = \frac{\ln(16.5)}{\ln(3)}$

d) $200(1.23^x) = 540$

Divide through by 200 to isolate $1.23^x$: $1.23^x = \frac{540}{200} = 2.7$

Take the natural logarithm of both sides: $\ln(1.23^x) = \ln(2.7)$

Using $\ln(a^b) = b \ln(a)$: $x \ln(1.23) = \ln(2.7)$

Solve for $x$: $x = \frac{\ln(2.7)}{\ln(1.23)}$

Let me calculate these values! math

Calculations for each part

a) 6^x = 78

x_a = math.log(78) / math.log(6)

b) (5.4)^x = 234

x_b = math.log(234) / math.log(5.4)

c) 8(3^x) = 132

x_c = math.log(16.5) / math.log(3)

d) 200(1.23^x) = 540

x_d = math.log(2.7) / math.log(1.23)

x_a, x_b, x_c, x_dHere are the solutions to the equations, rounded to three decimal places:

a) $x = 2.432$
b) $x = 3.235$
c) $x = 2.552$
d) $x = 4.798$

Would you like a detailed explanation of any specific part?

Follow-up Questions:

1. What is the significance of using natural logarithms to solve exponential equations?
2. How can these methods be applied to equations with different bases (e.g., base 10)?
3. What role do logarithmic properties play in simplifying these problems?
4. Can you identify the difference between natural logarithms ($\ln$) and logarithms of other bases?
5. How would the solution method change if the equations were in terms of a logarithmic function instead of an exponential one?

Tip:

Always double-check calculations when using logarithms by substituting the solution back into the original equation to verify its accuracy!