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4 years ago

What is the solution of log(t-3) = log(17-4t)?

Mathematics

2 answers:

Crazy boy [7]4 years ago

8 0

Answer:

t=4

Step-by-step explanation:

The given logarithmic equation is:

$\ log(t - 3) = \ log(17 - 4t)$

We want to solve for,

We take antilogarithm to get:

$t - 3 = 17 - 4t$

Combine similar terms to get:

$t + 4t = 17 + 3$

Simplify to obtain:

$5t = 20$

Divide both sides by 5

$t = \frac{20}{5} = 4$

NemiM [27]4 years ago

4 0

Answer:

$\mathrm{The\:solution\:is}$ :

$t=4$

Step-by-step explanation:

Given the expression

$log\left(t-3\right)\:=\:log\left(17-4t\right)$

$\mathrm{Apply\:log\:rule:\:\:If}\:\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\:\mathrm{then}\:f\left(x\right)=g\left(x\right)$

$t-3=17-4t$

$\mathrm{Add\:}3\mathrm{\:to\:both\:sides}$

$t-3+3=17-4t+3$

$t=-4t+20$

$\mathrm{Add\:}4t\mathrm{\:to\:both\:sides}$

$t+4t=-4t+20+4t$

$5t=20$

$\mathrm{Divide\:both\:sides\:by\:}5$

$\frac{5t}{5}=\frac{20}{5}$

$t=4$

Therefore, $\mathrm{the\:solution\:is}$ :

$t=4$

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3 years ago

Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.