Three Exponential Equations

Question

Could any of you guys assist me with the following questions. Thanks. I first tried moving everything to one side, but don't know how to continue.

Solve the following algebraically:

Question 1: $$5^{2x+1} + 25 = 5^{x+3} + 5^x$$

Answer is:

$x= -1 ,2$

Question 2: $$8^x +7\times2^{x+1} = 7\times4^x+8$$

Answer is:

$x= 0,1 , 2$

Question 3: $$3^{3x} -12\times9^x + 63\times 3^x = 4\times3^3$$

Answer is:

$x=1$

Thanks in advance.

Dr. Sonnhard Graubner · Accepted Answer · 2017-08-27T12:38:08.237

1

HINT: write your first equation as $$5\cdot (5^x)^2+25=125\cdot 5^x+5^x$$ and your second equation as $$(3^x)^3+14\cdot 2^x=7\cdot (2^x)^2+8$$ and the last one as $$(3^x)^3-12\cdot (3^x)^2+63\cdot 3^x=4\cdot 3^3$$

edited Aug 27 '17 at 12:38

answered Aug 27 '17 at 12:26

Dr. Sonnhard Graubner

95,283

In the first equation, where does the 125 come from? – Fuzzy Aug 27 '17 at 12:32
$5^{x+3}=5^3 \cdot 5^x = 125 \cdot 5^x$ – SvanN Aug 27 '17 at 12:33
Oh ok thanks heaps – Fuzzy Aug 27 '17 at 12:34
After writing the equations like that, what would be my next step – Fuzzy Aug 27 '17 at 12:37
set for example $t=5^x$ and solve the equation in $t$ – Dr. Sonnhard Graubner Aug 27 '17 at 12:39
So, the equation would look like $5\times(y)^2+25= 125y + y$ – Fuzzy Aug 27 '17 at 12:47
and the right-hand side $$5y^2+25=125\cdot y+y$$ – Dr. Sonnhard Graubner Aug 27 '17 at 12:51
Yes sorry, was meant to be 125. Now I factorize the equation? – Fuzzy Aug 27 '17 at 12:52
solve the quadratic equation $$5y^2-126y+25=0$$ for $y$ – Dr. Sonnhard Graubner Aug 27 '17 at 12:54
So, $5y^2-126y+25=0$ – Fuzzy Aug 27 '17 at 12:54
$(y-1)(y-125)$ right?
So, $5^x=1$ and $5^x=125$
– Fuzzy Aug 27 '17 at 12:55
this is ok and you can solve your equation – Dr. Sonnhard Graubner Aug 27 '17 at 12:57
So $x=0,3$ but the answer is supposed to be $x=-1,2$ – Fuzzy Aug 27 '17 at 12:59
the factorization is $$(5y-1)(y-25)=0$$ – Dr. Sonnhard Graubner Aug 27 '17 at 13:04
Oh yes, my mistake! Thanks heaps. – Fuzzy Aug 27 '17 at 13:05
you are welcome and a nice Sunday – Dr. Sonnhard Graubner Aug 27 '17 at 13:06

score 1 · Answer 2 · answered Aug 27 '17 at 12:26

1

Hint:

all these can be solved by substitution. In the first use $y=5^x$, in the second $y=2^x$ and do yourself for the last one.

answered Aug 27 '17 at 12:26

Emilio Novati

62,675

Three Exponential Equations

2 Answers2