using logarithms to solve exponential equations.
5^1+x = 2^1-x
I need exact numbers.
I did one on my own already. 5^x-1 = 9
5^x-1 = 9
log(5^x-1) = log9
(log5)(x-1) = log9
x-1 = (log9/log5)
x= (log9/log5)-1
x = 2.3652
Logarithm help - Joe, Friday, September 18, 2015 at 10:55pm
I really need help
Logarithm help - Savio, Friday, September 18, 2015 at 11:13pm
log(5^(1+x))= log(2^(1-x))
(1+x)(log 5)= (1-x)(log 2)
(x+1)/(x-1)=.4306765581
x+1= .4306765581x-.4306765581
At this point it's a simple algebraic solution to solve for x. :)
Logarithm help - Joe, Friday, September 18, 2015 at 11:27pm
The answer in the textbook says -0.398. I don't think the answer above gives me -0.398.
Logarithm help - Steve, Saturday, September 19, 2015 at 12:27am
(1+x)(log 5)= (1-x)(log 2)
(1+x)/(1-x) = log2/log5 = .4306765581
x = -0.39794
That doesn't make sense. How can you jump from 0.43067 to -0.39794?
come on - simple algebra:
(1+x)/(1-x) = .4306765581
1+x = .4306765581(1-x)
Think you can handle it now?
How about 1+x = 3(1-x)
To solve the equation 5^(1+x) = 2^(1-x) using logarithms, we can take the logarithm of both sides of the equation.
1. Take the logarithm of both sides:
log(5^(1+x)) = log(2^(1-x))
2. Apply the logarithmic property:
(1+x)log(5) = (1-x)log(2)
Now we have an equation with logarithms. We can simplify it further.
3. Distribute the logarithmic property:
log(5) + xlog(5) = log(2) - xlog(2)
4. Group the x terms together:
xlog(5) + xlog(2) = log(2) - log(5)
5. Combine the log terms using logarithmic properties:
x(log(5) + log(2)) = log(2/5)
6. Solve for x by dividing both sides by (log(5) + log(2)):
x = log(2/5) / (log(5) + log(2))
Now, you can plug in the values for log(2/5), log(5), and log(2) to get the numerical answer for x.
For the given equation, x = log(2/5) / (log(5) + log(2)) ≈ -0.39794.
To verify the result, you can substitute this approximate value back into the original equation and check if both sides are equal.