I see you've uploaded an image of a math problem related to logarithmic and exponential expressions. I will solve the tasks step by step. Here’s how I’ll proceed:
Problem Breakdown:
Simplify the expressions using logarithmic and exponential rules.
Convert all logarithms to natural logarithms (ln \ln ln
Identify the domain for x x x
Let me start with each subproblem:
a) exp ( 3 ln 4 ) \sqrt{\exp(3 \ln 4)} exp ( 3 ln 4 )
\exp(3 \ln 4) = 4^3 \quad \text{(since \exp(a \ln b) = b^a)}
exp ( 3 ln 4 ) = 4 3 = 8 \sqrt{\exp(3 \ln 4)} = \sqrt{4^3} = 8 exp ( 3 ln 4 ) = 4 3 = 8
Answer : 8 8 8
b) exp ( ( x + 2 ) 2 − 4 ) e x 3 \sqrt[3]{\frac{\exp((x+2)^2 - 4)}{e^x}} 3 e x e x p (( x + 2 ) 2 − 4 )
exp ( ( x + 2 ) 2 − 4 ) = e ( x + 2 ) 2 ⋅ e − 4 \exp((x+2)^2 - 4) = e^{(x+2)^2} \cdot e^{-4} exp (( x + 2 ) 2 − 4 ) = e ( x + 2 ) 2 ⋅ e − 4 exp ( ( x + 2 ) 2 − 4 ) e x = e ( x + 2 ) 2 − x − 4 \frac{\exp((x+2)^2 - 4)}{e^x} = e^{(x+2)^2 - x - 4} e x e x p (( x + 2 ) 2 − 4 ) = e ( x + 2 ) 2 − x − 4 exp ( ( x + 2 ) 2 − 4 ) e x 3 = e ( x + 2 ) 2 − x − 4 3 \sqrt[3]{\frac{\exp((x+2)^2 - 4)}{e^x}} = e^{\frac{(x+2)^2 - x - 4}{3}} 3 e x e x p (( x + 2 ) 2 − 4 ) = e 3 ( x + 2 ) 2 − x − 4
Answer : e ( x + 2 ) 2 − x − 4 3 e^{\frac{(x+2)^2 - x - 4}{3}} e 3 ( x + 2 ) 2 − x − 4
c) log 2 ( e x ) \log_2(\sqrt{e^x}) log 2 ( e x )
e x = e x / 2 \sqrt{e^x} = e^{x/2} e x = e x /2 log 2 ( e x / 2 ) = x 2 ⋅ log 2 ( e ) \log_2(e^{x/2}) = \frac{x}{2} \cdot \log_2(e) log 2 ( e x /2 ) = 2 x ⋅ log 2 ( e )
Answer : x 2 log 2 ( e ) \frac{x}{2} \log_2(e) 2 x log 2 ( e )
d) 1 2 log 2 ( 4 e 2 ) − ( ln 2 ) − 1 \frac{1}{2} \log_2(4e^2) - (\ln 2)^{-1} 2 1 log 2 ( 4 e 2 ) − ( ln 2 ) − 1
log 2 ( 4 e 2 ) = log 2 ( 4 ) + log 2 ( e 2 ) = 2 + 2 log 2 ( e ) \log_2(4e^2) = \log_2(4) + \log_2(e^2) = 2 + 2\log_2(e) log 2 ( 4 e 2 ) = log 2 ( 4 ) + log 2 ( e 2 ) = 2 + 2 log 2 ( e ) 1 2 log 2 ( 4 e 2 ) = 1 2 ( 2 + 2 log 2 ( e ) ) = 1 + log 2 ( e ) \frac{1}{2} \log_2(4e^2) = \frac{1}{2}(2 + 2\log_2(e)) = 1 + \log_2(e) 2 1 log 2 ( 4 e 2 ) = 2 1 ( 2 + 2 log 2 ( e )) = 1 + log 2 ( e ) ( ln 2 ) − 1 = 1 ln 2 = log 2 ( e ) (\ln 2)^{-1} = \frac{1}{\ln 2} = \log_2(e) ( ln 2 ) − 1 = l n 2 1 = log 2 ( e ) 1 2 log 2 ( 4 e 2 ) − ( ln 2 ) − 1 = 1 + log 2 ( e ) − log 2 ( e ) = 1 \frac{1}{2} \log_2(4e^2) - (\ln 2)^{-1} = 1 + \log_2(e) - \log_2(e) = 1 2 1 log 2 ( 4 e 2 ) − ( ln 2 ) − 1 = 1 + log 2 ( e ) − log 2 ( e ) = 1
Answer : 1 1 1
e) ln ( x 2 3 ) − ln ( x 4 3 ) \ln\left(\frac{x^2}{3}\right) - \ln\left(\sqrt[3]{\frac{x}{4}}\right) ln ( 3 x 2 ) − ln ( 3 4 x )
ln ( x 2 3 ) − ln ( x 4 3 ) = ln ( x 2 3 ) − 1 3 ln ( x 4 ) \ln\left(\frac{x^2}{3}\right) - \ln\left(\sqrt[3]{\frac{x}{4}}\right) = \ln\left(\frac{x^2}{3}\right) - \frac{1}{3} \ln\left(\frac{x}{4}\right) ln ( 3 x 2 ) − ln ( 3 4 x ) = ln ( 3 x 2 ) − 3 1 ln ( 4 x ) = ln ( x 2 3 ) − 1 3 ( ln x − ln 4 ) = \ln\left(\frac{x^2}{3}\right) - \frac{1}{3} (\ln x - \ln 4) = ln ( 3 x 2 ) − 3 1 ( ln x − ln 4 ) = ln x 2 − ln 3 − 1 3 ln x + 1 3 ln 4 = \ln x^2 - \ln 3 - \frac{1}{3} \ln x + \frac{1}{3} \ln 4 = ln x 2 − ln 3 − 3 1 ln x + 3 1 ln 4 = 5 3 ln x − ln 3 + 1 3 ln 4 = \frac{5}{3} \ln x - \ln 3 + \frac{1}{3} \ln 4 = 3 5 ln x − ln 3 + 3 1 ln 4
Answer : 5 3 ln x − ln 3 + 1 3 ln 4 \frac{5}{3} \ln x - \ln 3 + \frac{1}{3} \ln 4 3 5 ln x − ln 3 + 3 1 ln 4
f) ln ( 2 x ) + ln ( 3 x ) − ln ( x 2 ) − ln ( 6 ) \ln(2x) + \ln(3x) - \ln(x^2) - \ln(6) ln ( 2 x ) + ln ( 3 x ) − ln ( x 2 ) − ln ( 6 )
Using the logarithmic property ln a + ln b = ln ( a b ) \ln a + \ln b = \ln(ab) ln a + ln b = ln ( ab ) ln ( 2 x ) + ln ( 3 x ) = ln ( 6 x 2 ) \ln(2x) + \ln(3x) = \ln(6x^2) ln ( 2 x ) + ln ( 3 x ) = ln ( 6 x 2 ) ln ( 6 x 2 ) − ln ( x 2 ) − ln ( 6 ) = ln ( 6 x 2 x 2 ⋅ 6 ) \ln(6x^2) - \ln(x^2) - \ln(6) = \ln\left(\frac{6x^2}{x^2 \cdot 6}\right) ln ( 6 x 2 ) − ln ( x 2 ) − ln ( 6 ) = ln ( x 2 ⋅ 6 6 x 2 ) = ln ( 1 ) = 0 = \ln(1) = 0 = ln ( 1 ) = 0
Answer : 0 0 0
g) log 10 ( 10 x 10 ) \log_{10}(10x^{10}) log 10 ( 10 x 10 )
log 10 ( 10 x 10 ) = log 10 ( 10 ) + log 10 ( x 10 ) \log_{10}(10x^{10}) = \log_{10}(10) + \log_{10}(x^{10}) log 10 ( 10 x 10 ) = log 10 ( 10 ) + log 10 ( x 10 ) = 1 + 10 log 10 ( x ) = 1 + 10\log_{10}(x) = 1 + 10 log 10 ( x )
Answer : 1 + 10 log 10 ( x ) 1 + 10\log_{10}(x) 1 + 10 log 10 ( x )
h) log 3 ( x 2 − 4 ) − log 3 ( 3 ( x − 2 ) ) \log_3(x^2 - 4) - \log_3(3(x-2)) log 3 ( x 2 − 4 ) − log 3 ( 3 ( x − 2 ))
Using log a ( m ) − log a ( n ) = log a ( m n ) \log_a(m) - \log_a(n) = \log_a\left(\frac{m}{n}\right) log a ( m ) − log a ( n ) = log a ( n m ) log 3 ( x 2 − 4 ) − log 3 ( 3 ( x − 2 ) ) = log 3 ( x 2 − 4 3 ( x − 2 ) ) \log_3(x^2 - 4) - \log_3(3(x-2)) = \log_3\left(\frac{x^2 - 4}{3(x-2)}\right) log 3 ( x 2 − 4 ) − log 3 ( 3 ( x − 2 )) = log 3 ( 3 ( x − 2 ) x 2 − 4 ) x 2 − 4 = ( x − 2 ) ( x + 2 ) x^2 - 4 = (x-2)(x+2) x 2 − 4 = ( x − 2 ) ( x + 2 ) log 3 ( ( x − 2 ) ( x + 2 ) 3 ( x − 2 ) ) = log 3 ( x + 2 3 ) \log_3\left(\frac{(x-2)(x+2)}{3(x-2)}\right) = \log_3\left(\frac{x+2}{3}\right) log 3 ( 3 ( x − 2 ) ( x − 2 ) ( x + 2 ) ) = log 3 ( 3 x + 2 )
Answer : log 3 ( x + 2 3 ) \log_3\left(\frac{x+2}{3}\right) log 3 ( 3 x + 2 )
Let me know if you'd like detailed explanations for specific steps or further clarifications!
Here are 5 related questions for exploration:
How are logarithmic properties derived from exponential functions?
What are the applications of logarithms in solving real-world problems?
How does the base of a logarithm influence its calculation?
Can you generalize rules for logarithmic differentiation?
What’s the importance of identifying domains for expressions involving logarithms?
Tip : Always double-check if the argument of a logarithm is positive, as logarithms are undefined for non-positive arguments.