SOLUTION: Solve the exponential equation. Round the result to 3 decimal places. 1. {{{2^(x+1)=3^(2x)}}} 2. {{{5^(2x+1)=2^(4x-3)}}} Solve the Logarithmic equation. Round the result to 3

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve the exponential equation. Round the result to 3 decimal places. 1. {{{2^(x+1)=3^(2x)}}} 2. {{{5^(2x+1)=2^(4x-3)}}} Solve the Logarithmic equation. Round the result to 3       Log On


   

Question 34259This question is from textbook McDougal Littell Algebra 2
: Solve the exponential equation. Round the result to 3 decimal places.
1. 2%5E%28x%2B1%29=3%5E%282x%29
2. 5%5E%282x%2B1%29=2%5E%284x-3%29
Solve the Logarithmic equation. Round the result to 3 decimal places.
1.+log+2+%28x%2B1%29=+log+4+%282x-3%29
2.+log+%28%28x-4%29%29=+log+100+%28x%2B3%29
These problems come from pg 85 [Lesson 8.6, Practice C] of Mcdougal Littell's Algebra 2 Chpt. 8 Resource book
Help me A.S.A.P with shown work This question is from textbook McDougal Littell Algebra 2

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
#1 Take the log of both sides to get:
(x+1)(log2)=(2x)(log3)
xlog2+ log2 = 2x(log3)
x(log2-2log3)=-(log2)
x(-0.653212514...)=-0.30109996...
x=0.460845421...
#2 I'll leave this to you as it is similar to #1
#3[log(x+1)/log2]=[log(2x-3)/log4]
Note that log4 = 2log2, So we can write both fractions over the same denominator. Then equating the numerators you get:
[2(log(x+1)]=log(2x-3)
Then log(x+1)^2= log(2x-3)
Taking the anti-log get:
(x+1)^2=2x-3
x^2+2x+1=2x-3
x^2+4=0
x=2i or x=-2i
The x values are imaginary; the problem has no real solution.
Cheers,
Stan H.