5^x =4^(x+1)?

Question

its logarithmic

Answer 1

take "ln's" of both sides:

ln (5^x) = ln (4^[x+1])

then

x * ln 5 = [x+1] * ln 4

xln5 = xln4 + ln4

x = (ln 4) / (ln 5 - ln 4)

Answer 2

5^x = 4^(x + 1)

Get the (x + 1)th root of both sides:
5^[x/(x + 1)] = 4

Convert into logarithmic form:
log_5 (4) = x/(x + 1)

Multiply by x + 1
(x + 1) (log_5 (4)) = x

distribute
log_5 (4) x + log_5 (4) = x

transpose x and log_5 (4)
log_5 (4) x - x = - log_5 (4)

Factor out x
x[log_5 (4) - 1] = - log_5 (4)

Multiply by -1
x[1 - log_5 (4)] = log_5 (4)

Divide
x = log_5 (4)/[1 - log_5 (4)]

Hope this helps^_^
^_^

Answer 3

log both sides:
log(5^x) = log (4^(x+1))

Use the exponent property:
x(log 5) = (x+1)(log 4)

Distribute:
x(log 5) = x(log 4) + log 4

Move the x's to one side:
x(log 5) - x(log 4) = log 4

Factor out x:
x(log 5 - log 4) = log 4

Isolate the x:
x = (log 4) / (log 5 - log 4)

Answer 4

5^x = 4^(x + 1)
xln5 = (x + 1)ln4
xln5 = xln4 + ln4
xln5 - xln4 = ln4
x(ln5 - ln4) = ln4
x = (ln4)/(ln5 - ln4)
x = about 6.212567

ANS : x = about 6.21

Answer 5

Yes -- it is logarithmic.

so:

LN(5^x) = LN(4^(X+1))

x*LN(5) = (x+1)*LN(4)

x*LN(5) = x*LN(4) + LN(4)

x*(LN(5)-LN(4)) = LN(4)

X*LN(5/4) = LN(4)

X = LN(4)/LN(5/4)

Answer 6

5^x=4^(x+1)
5^x=4^x*4
5^x/4^x=4
(5/4)^x=4
xln(5/4)=ln4
x=ln4/ln(5/4)=0.602/0.096
x=6.27

Answer 7

ya can say

x comes to be 6.21256

Answer 8

i am not sure!!
y=5^x=4^(x+1)
so logy(base5)=logy(base4)-1
logy(base4)-logy(base5)=1
1/log4(basey)-1/log5(basey)=1
so {log5(basey)-log4(basey)}
/{log4(basey)*log5(basey)}=1
log(5/4)base(y)/
{log4(basey)*log5(basey)}=1
log (5/4)base4=log5(baesy)
(1/8)log5(base2)=log5(baesy)
8=logy(base2)
so y=2^8 and X=8*log2(base5)