Question

i need to solve for x:
\[x+\lg(1+2^{x})=xlg5+\lg6\]

Answer 1

@amistre64

Answer 2

x+g6l2x+2xg6l2x+g6l

Answer 3

maybe:

\[x+\lg(1+2^{x})=\lg5^{x}+\lg6\]

Answer 4

then:

\[x+\lg(1+2^{x})=\lg(5^{x}*6)\]
but what to do next? @phi?

Answer 5

what base is lg ?

Answer 6

\[x=\lg(\frac{ 6*5^{x} }{1+2^{x} })\]

Answer 7

lg is \[\log _{10}\]

Answer 8

@beneemi I am sorry, can you explain what you have done?

Answer 9

a friend did this answer i have know clue sorry

Answer 10

oh, it doesnt make sense to me:/

Answer 11

you could undo the log
10^x = 6 * 5^x / (1+2^x)
and write
10^x as (2*5)^x = 2^x * 5^x

Answer 12

thank you, @phi

Answer 13

I get
\[ 2^x \cdot 5^x = \frac{6 \cdot 5^x}{(1+2^x)} \]
divide both sides by 5^x
\[ 2^x = \frac{6 }{(1+2^x)} \]
multiply through by 1+2^x
\[ 2^{2x} +2^x -6 =0 \]
or
\[ \left(2^x\right)^2 + 2^x - 6 =0 \]
this is a quadratic with roots -3 and 2

Answer 14

we can discard the -3. we are left with
2^x = 2
and
x= 1