Answer:
x = -2/ 3
Step-by-step explanation:
in order to cancel out the logs they should have common bases
[tex] \mathsf{ log_{5}(3x + 5) = 2 log_{25}(1 - 3x) } [/tex]
we can write 25 as 5²
[tex]\mathsf{ \implies log_{5}(3x + 5) = 2 \: log_{ {5}^{2} }(1 - 3x) } [/tex]
we know that the reciprocal of the exponents of the bases are multiplied to the log
[tex]\mathsf{ \implies log_{5}(3x + 5) = \frac{1}{\cancel{2} } \times \cancel{2} \: log_{ 5}(1 - 3x) } [/tex]
and now since the logs have common bases
[tex]\mathsf{ \implies \cancel { log_{5}}(3x + 5) = \cancel{ log_{ 5}}(1 - 3x) } [/tex]
we're left with
[tex]\mathsf{ \implies 3x + 5 = 1 - 3x} [/tex]
[tex]\mathsf{ \implies 9x = -4} [/tex]
x = -2/ 3
