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OK......I guess we have 

2^(t + 3)   = 7^(t - 1)    ????    if so........take the log of both sides

log2^(t +3)  = log7^(t - 1)      and we can write

(t +3) log2  = (t - 1) log 7     simplify

t log2  + 3log2  = tlog7 - log 7

3log2 + log7  = tlog7 - t log2

3log2 + log 7  = t [ log 7 - log2]       divide both sides by log 7 - log2

[3log2 + log 7 ] / [ log7 - log2]  = t  = about 3.2132

12  24  36  48  60 are multiples of 12

        16 32 48               are multiples of 16

   ....the answer is 48

i think u misunderstood...those are raised to the power............it has somethig to do with log brother.

A BIG stomach ache and an even bigger BM.

-451146908858814354332915157103023924108293663848310699255997391822

Here's a graphical solution :  https://www.desmos.com/calculator/hvjfai7rfa

The solution occurs at about t = 0.91

look at the symble if there isent one than dont solve it and tell the person who gave it to you it cant be solved becuse theres no symble

Wow,   this poster could REALLY use some instruction in English.   Misspellings, improper punctuation of possessive nouns etc....   Pathetic....and really trying (but failing!) to be funny.

Look under the 'Widgets' tab from the home page....

Using quadratic formula?

what is the GCF of 42x and 12xy tell me the answer not explain it plz need it thx

whats the answer for this equation -3-9a-4-5a=7

-61=d+(-18)  add 18 to both sides

-43  = d

(2x+1)*(6x-1)  =  12x^2 + 6x - 2x - 1   =  12x^2 + 4x - 1

We don't have enough info to solve this......your equation is correct, but s and a may - or may not - have integer solutons with just one equation..........we need to generate another equation  involving s and a, somehow

grow up and get a job

hlp plz

There's  nothing to "solve" ......we can "expand" this as......

(x - 4)^2  = (x - 4) ( x - 4)  =  x^2 - 8x + 16

find the value of x      3x+5 =7x+21

Use a combination of graphical and numerical methods:

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I really like your explanation Chris :)

show that 1+i is a solution for the equation x^2-2x+2=0

\((1+i)^2 =1^2+2i+i^2=1+2i-1=2i\\ -2(1+i)=-2-2i\\ so\\ (1+i)^2-2(1+i)+2\\ =2i-2-2i+2=0\\ \therefore\\ 1+i \quad \mbox{is a solution to }\;\;x^2-2x+2=0\)

If the graph of y = |x| is translated so that the point (1, 1) is moved to (1, 4), what is the equation of the new graph?

If you think about those 2 points, the new one is exactly 3 units directly above the old one so it is just the old y value + 3 more.

y=|x|+3