Answer:
A.)
Step-by-step explanation:
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
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You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
nsweres el dodeel s Step-by-step explanationen
By doing by over 2 it will equal yeah
10.5 hours
let t be time then distance travelled by both is
Mr Burges ⇔ d = 50t ( distance = speed × time )
Mrs Burges ⇒ d = 70(t - 3)
At the point of overtaking both will have travelled the same distance
equating their distance and solving for t
70(t - 3) = 50t
70t - 210 = 50t ( subtract 50t from both sides )
20t - 210 = 0 ( add 210 to both sides )
20t = 210 ( divide both sides by 20 )
t = 10.5 hours
