solve 5^y=4^y+3.Round your answer to the nearest tenth.
I will assume you meant
5^y = 4^(y+3) , or else you would have a very messy equation
take log of both sides
log 5^y = log 4^(y+3)
y log5 = (y+3)log4
y log5 = y log4 + 3 log4
ylog5 - ylog4 = 3log4
y(log5 - log4) = 3log4
y = 3log4/(log5-log4)
I will let you do the button-pushing, (I got 18.6)
I got 6.46
I did it a second time and got 1.98
I get 18.6
95 - Did you see my repair of the sign error I made in your hyperbola problem?
note - log 5 - log 4 = log 1.25
Damon- yes, thank you
How did you get 18.6? on my calculator I typed in 3*log(4)/log(5)-log(4).Am I doing this wrong? solving it wrong?
I see some missing parentheses in the denominator. Or, rather, I don't see them.
To solve the equation 5^y = 4^y + 3, we need to isolate the variable y.
Step 1: Start by subtracting 4^y from both sides of the equation to move all terms involving y to one side:
5^y - 4^y = 3
Step 2: Notice that both sides of the equation have the same base (5 and 4). We can use the fact that a^x - b^x can be factored as (a - b)(a^(x-1) + a^(x-2)b + ... + b^(x-1)). Applying this rule, we can factor out a common term:
(5 - 4)(5^(y-1) + 5^(y-2)4 + ... + 4^(y-1)) = 3
Simplifying further:
5^(y-1) + 5^(y-2)4 + ... + 4^(y-1) = 3
Step 3: Now we need to solve the equation involving the variable y. Unfortunately, there is no algebraic method to solve this equation directly. We will have to use numerical methods.
One numerical method is the trial-and-error method. We can try different values for y, substitute them into the equation, and see which values make the equation true. Since you want the answer rounded to the nearest tenth, we will start by trying values close to 0. We can try y = -1, 0, 1, and so on.
Let's use a scientific calculator or an online calculator to evaluate the left side of the equation for each value of y:
For y = -1: 5^(-1) + 5^(-2)4 = 0.32 (approximately)
For y = 0: 5^0 + 5^0(4) = 5 (approximately)
For y = 1: 5^1 + 5^04 = 6.2 (approximately)
By trial and error, we can continue testing values of y until we find the one that makes the equation true. Continue this process until you get closer to a specific value. Then, use that value as a starting point in an iterative method, such as the Newton-Raphson method or the bisection method, to find a more accurate solution.
Once you have found a solution, round it to the nearest tenth.
Please note that without further context or constraints, this is a general explanation of how to solve the equation.