SOLUTION: Please help me solve this problem
3^x+4=5^x-1
(x+4)log3=(x-1)log5
xlog3+4log3=xlog5-1log5
4log3+1log5=xlog3-xlog5
x=4*0.47712+1*0.69897/0.69897-0.47712
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Question 145713: Please help me solve this problem
3^x+4=5^x-1
(x+4)log3=(x-1)log5
xlog3+4log3=xlog5-1log5
4log3+1log5=xlog3-xlog5
x=4*0.47712+1*0.69897/0.69897-0.47712
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
There is a small error in step 4:
3^x+4=5^x-1
(x+4)log3=(x-1)log5
xlog3+4log3=xlog5-1log5
4log3+1log5=xlog3-xlog5 SHOULD BE 4log3 + 1 log 5= xlog5 - x log 3
Next, factor the x from the right side of the equation:
4log3 + 1 log 5= x(log5 - log 3)
Last, divide both sides by (log5 - log3)
Can you use a calculator to calculate this?? I get x=11.75.
For additional help please see my website by clicking on my tutor name "rapaljer" anywhere in algebra.com. Look for the second link on my homepage which is "Basic, Intermediate, and College Algebra: One Step at a Time." Choose "College Algebra", then "Chapter 4 Logarithms." This lesson is "Section 4.04 Solving Exponential and Logarithmic Equations." My own students used this book for about 20 years, and they unanimously proclaimed that it was easier to understand than any other book they ever used!! I hope you like it. See also the "MATH IN LIVING COLOR" pages that are linked to these sections.
R^2
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